Good evening everyone. Today we are in the fourth of six sessions and we have quite a bit to cover. There are two general topics that we're going to be covering. So the fourth session refers to axially loaded members. Essentially these are columns so they're not only axially loaded they are subjected to compression and bending. So there are two general topics that we're going to cover today. We're going to cover interaction diagrams that you might remember from reinforced concrete design and we will also cover slenderness effects. So with that I'll just move right into the material because like I said we have quite a bit of material to cover tonight. So the idea as I mentioned just briefly a few seconds ago, these are the major outcomes that I intend to cover. Understand what a load moment interaction diagram is. Be able to, so the second outcome is be able to calculate the strength of compression members under pure axial load. What are the different factors that affect interaction diagrams, PM interaction diagrams. And also review or gain an understanding of what ACI code provisions exist on slender compression members. So those are the general outcomes that we'll be covering today. So first we're going to talk about interaction diagrams. The idea of an interaction diagram is that really represents the strength envelope of a member that is subjected to compression under the simultaneous application of moment. So an interaction diagram represents really the strength envelope of a given cross-section. It is important to remember that it constitutes an ultimate condition. So that means that the surface or the curve that is drawn for the interaction diagrams is a strength, it represents strength not service condition. The construction of it as we will see in a few slides is based on assumed material strengths and some simplifications of the material models. And what's important to remember as well is that in the PCI design handbook there's a series of resources to construct interaction diagrams. Section 510.1 has a series of steps that will guide you through the construction of interaction diagrams. And figure 510.1 illustrates representative points of what are some key points in the construction of an interaction diagram. So I would encourage you to visit both the section and the figure that are listed in this slide. And I'll just because of time I will only be able to list the highlights. Some of the simplifications that are employed in construction of interaction diagrams are listed here. The first is should be common hopefully to most of you in that the concrete stresses in the cross section at the strength limit state is replaced and represented using the Whitney stress block. So it is replaced by a stress block of constant magnitude. The uniform stresses that are applied in the compression zone of a member equal 0.85 f prime C. The depth of the block is referred to as A and it is a constant beta 1 times C, the neutral axis depth. The other material simplification is that mild reinforcing steel is considered elastoperfectly plastic. So when we reach yield that level is maintained at large deformation so steel does not strain harden. And the last simplification that I'm listing here is that prestressing steel behaves nonlinearly at ultimate and its stresses can be determined using the design aid that is in section 15.2.3 in the PCI design handbook. So these are very very quick simplifications that I wanted to make you aware of interaction diagrams. And just as an introductory material I also wanted to illustrate what's going to be quite useful what I often call to as a statically equivalent condition. So most of the time engineers recognize that columns are subjected to the simultaneous action of an axial force and a bending moment about some reference axis in the cross section, in this case the y-axis. So this actual loading can be replaced by a static replacement would be to move the axial force to a certain eccentricity. And the static equilibrium or the static equivalency is such that this eccentricity has to be calculated in such a way that it represents or equates with the moment that's acting on the section. So in other words the moment about the y-axis has to be equal to the axial force times this eccentricity. So these two conditions are equivalent. So in many cases we'll be referring to an axial force acting at an eccentricity, in this case EX, or in other cases we'll be referring to an axial force and moment. These two conditions are exactly the same from a static point of view. And where do these moments arise? So you can see in this slide for precast pre-stressed concrete buildings there are actual eccentricities. So for example, I have a few examples here in this slide. One example here is a column in a perhaps in a parking garage with two corbels in which we have a long span on the left-hand side, a short span on the right-hand side. So because of that the axial forces that are being generated into the column or as a reaction to the corbel directly have different magnitudes. In this case P2 is less than P1. If the eccentricities are the same we know that there's an unbalance in moments. So one could conceive, one could think of well the total axial force in the column is P1 plus P2 and the resulting moment is the difference between P1 and P2 times the eccentricity. So the difference in load is what's causing the moment. There are other cases where you know a horizontal force such as that due to wind or earthquake actually generates a bending moment say at the base of this column modeled as a cantilever. So there's axial force from the floor loads acting in the column and a horizontal force that generates a moment. Or there could be just simply a one-sided corbel with an axial force coming from the floor system acting at that eccentricity that also generates a P here and a moment. So this is similar to the simplification I just mentioned earlier. One thing worth noting is that because of light load patterning within the floor or throughout the building type, different eccentricities may be caused and generating moments. In the case of a parking rudge which is the example that we've been solving here in this class so far, light load patterning is not allowed so that wouldn't apply to that particular condition. This would be for an office building or some other building of a different use. So in the next few slides we're going to go through the different points in the interaction diagram. So I wanted to get started with the first poll of this evening by asking you and asking Sherry to launch it. I wanted to to ask whether you remembered the steps that are used to construct a PM interaction diagram. All right everyone I will be launching the poll now. Please answer the poll on your screen. Thank you so much for doing so and I'll leave it open for a few more minutes so everyone can partake. Okay I'm going to close the poll now and Sergio it was 50-50. Okay thank you. All right so I will spend some time as much as I can talking about the different points. So the idea in an interaction diagram is to construct different load combinations that combined represent failure of a cross-section. So the basic premise is that you have a given cross-section such as shown here in this slide and what's shown here is one of the key points in an interaction diagram is you know asking the question or determined by by saying well what is the maximum axial force that a column cross-section can take prior to failure. So if that is the case this red dot here in the center of the column represents the point of application of a concentrically applied axial force and we have a reinforcing bar arrangement. Each layer of reinforcement is labeled through numbers 1, 2, and 3. So layer 1 has three bars, layer 2 two bars, and so on. The total area of seal is labeled as AST. So for failure of the concrete section to occur, similar to reinforced concrete design, the strains have compressive strains in the concrete have to reach a value of 0.003 inches per inch. So that is an assumption that that crushing of concrete occurs at a strain of 0.003 and that's commonly used for concrete as you might remember. So the capacity in concentric axial compression is simply calculated using this strain profile and recognizing that at that level what if we use the replacement of an equivalent stress block the stresses in the concrete are going to be at a magnitude of 0.85 f prime C acting over the cross-sectional area of the concrete so that's 0.85 f prime C over the entire cross-section minus the area that's being used up by the steel so the area occupied by the steel. That's the total compression force in the in the concrete without counting how much compression force the steel bars resist. So the steel bars resist the value of each of the layers of steel A as 1, 2, and 3. All of them are yield as long as this strain exceeds the yield strain of the reinforcement which for grade 60 reinforcement that is the case. So we add the contribution of the steel to the compressive strength and we come up with by summation of vertical forces P equals 0 we get to the total the nominal capacity of the section without any eccentricity. So if this section were made up of not only mild reinforcement but also of pre-stressing strand we would need to account in the for this instead of using FY we would have to use FPS which would be the stress in the pre-stress reinforcement at failure. I will comment more on that in a subsequent slide. Okay so how sorry going forward so one of the figures or part of the figure in figure 5.10.1 this is part A shows you how to calculate different points in the interaction diagram. So for example this is for a condition where the axial force and notice that the the figure here slightly different from what I showed you in the previous slide the axial force is acting at an eccentricity about some reference line so the cross-section looks like the cross-section of a part of a wall say with with some you know this irregular T like section where the axial force is acting somewhere along here at an eccentricity E. So it really this section is being acted upon by an axial force in a moment. So the what what is important to recognize is that this moment will will produce let's say if you think of the section as the top being in compression the bottom in tension so the top part here is going to be at this crushing strain of the concrete for the failure condition to be reached. So if we know this and we actually assume a location so for this particular condition I should say pardon me let's assume that the neutral axis depth is at a distance C from the compression face. If we know the distance C and this value of 0.003 we can then determine any other strains of the components in the cross-section labeled as epsilon is prime epsilon PS prime epsilon PS and epsilon S here. So notice that regions here above the line this horizontal line are compression and regions below are in tension. The shaded area here is the area of concrete that is subjected to these constant stresses of magnitude 0.85 f prime C that react against this applied force acting at an eccentricity E. So that is these stresses are acting over this irregular shape and the centroid of that irregular shape is at this X designated as Y prime from the compression face. Similarly all the other component forces this area of pre-stressing steel APS prime pardon me this area of mild compression reinforcement AS prime is at a distance D prime. This area of compression steel pre-stressing compression reinforcement is a DP prime and all the other depths are measured as well from the compression phase DPD and H is the depth of the entire cross-section. So each of the component forces here we illustrate AS FS prime. AS prime FS prime is also a compression force in the mild reinforcement that reacts against PN. Notice that APS prime and FPS prime are acting towards the left meaning tension in the internal tension and this is because it is very likely that this pre-stressing steel after loading it's still going to be in tension because it was pre-stressed prior to application of the load. Again I will mention more of this later on. Then the other two tension forces are APS FPS and AS FS. You will notice that all of these values of strain can be computed using this linear relation using similar triangles. An example is given here for epsilon S prime. You can follow from similar triangles and find epsilon S prime that will then allow us to determine FS prime. Similarly epsilon S can be found that way and the other strains in the pre-stressing strand are found by adding from similar triangles from the loading and adding the pre-stressing the effective pre-stressing strain to each of these components. So for example for epsilon prime PS we have the strain induced by effective pre-stressing and then we subtract the to that we subtract this term that comes up from epsilon PS prime because it's in the compression phase or on the compression side of this column or wall. Whereas in the other equation here we add it because this effective pre-stressing strain we add the strain induced by loading on this section. So with these strains we can then calculate FPS and FPS prime and then we can compute internal forces in the cross-section. So knowing epsilon PS prime and epsilon PS we use the design 8.15.2.3 in the PCI design handbook and let's say this is epsilon PS we move vertically here let's say it's grades 270 reinforcement then we can read FPS from this graph shown here. Notice the assumption of elasto-perfectly plastic mild reinforcement is shown here so you multiply the strains times modulus of steel but it cannot exceed the yield stress so that's what this is a limit for that reinforcement and again I just mentioned how to calculate FPS and FPS prime. And finally the last step in calculating this point this particular point in the PM diagram would be to sum all of these internal forces that are taking place inside the cross-section so in this case we would sum it looks like horizontal force equilibrium but it's actually a vertical force equilibrium it's just the graph is rotated 90 degrees so we sum up all forces in the axial direction and you'll see the signs corresponds to the direction of the arrows with positive being to the right and moments excuse me clockwise moments are calculated using all the terms that produce a moment about this reference line about which the eccentricity is measured so PN times E gives us a reference line by which we should calculate moments and it is important to recognize that if we try to calculate moments about any other reference line in this case it's the center of gravity of the section the resulting moment will change because there's an effective axial force so this line is quite important you have to be consistent about which line you calculate moments as noted here in the bottom of the slide so hopefully you'll be able to follow the just these equations by looking at the graph here I wanted to go into detail of what's in that figure in the PCI design handbook so that you can look at other parts of that figure and and understand what's going on so I'm going to now go through a series of slides that demonstrate characteristics of an interaction diagram so this is the shape the general shape of an interaction diagram a point such as the one we just calculated would graph somewhere here let me just call that point I it would correspond to the moment and axial force that we compute previously here so PN and MN will give me that point in the interaction diagram and that's just for a given eccentricity EX like I'm just calling it EX here so this is what's shown here an axial force and a moment you know plotted point by point the point of pure compression that we discussed earlier this recent point might plot here and similarly you determine other points along the interaction diagram at different eccentricities please note that these radial lines that start from the origin and have an is an slope equal to 1 divided by the eccentricity are referred to as constant lines or lines of constant eccentricity they I often try to call them these lines of equal eccentricity imagine that you have zero axial force and you increase the axial force at a constant eccentricity until you reach failure of the section that's what this loading path would represent until you reach the capacity of the section that's represented by this interaction diagram another important point here is what's referred to as a balance failure balance failure is the condition when the steel reinforcement in the extreme tension fiber yields at the same time the concrete crushes that's balanced condition and above those that line are compression controlled failures and below it are either transition or tension controlled failures and you might recall from reinforced concrete design what those mean in the box here's what the explanation of compression control failures and transition or tension control failures are let me just read for compression control control failures is where crushing our concrete occurs prior to the extreme layer of steel yielding in tension another example so what is the effect of eccentricity on the axial strength recall that we can represent moment by eccentric axial force so let's say we have point I and we have a different point J of the same column where the axial force is applied at a larger eccentricity EXJ so if this eccentricity is larger it would be equivalent to loading at a shallower slope here and then one over EXJ and as you see the point where that line crosses the interaction diagram corresponds to a lower axial force so in effect this point or this failure point represents a failure where because of the larger moment being applied to the column the column section cannot take as much axial force as it did when the eccentricity was at EXI so added eccentricity has the effect of reducing the axial capacity in general until you get to to the balanced condition so that's an important observation in interaction diagram so let me just go through some calculations I will for illustration purposes I'm just going to use an 18 by 18 inch cross-section that has six number nine bars distributed this way three on the top three on the bottom but the bending is going to be about the y-axis and the x-axis so we're going to construct interaction diagrams nominal and design interaction diagrams about both axes independently and we're going to discuss the characteristics this is the objective these are the concrete and and steel mild steel reinforcing properties and I created a spreadsheet to perform the calculations and I will show you one calculation or one point in the spreadsheet I used so the result I'm just going to jump ahead and give you the result so given the cross-section 18 by 18 with six number nines if bending is being applied about the y-axis you would get this green line to represent the interaction diagram for bending about the y-axis for bending about the x-axis you would get this blue line notice that the axial compression concentric axial compression point is exactly the same in both because there's no difference in the cross-section also the concentric axial tension that represents the maximum tension force a column can take which is just the area of reinforcement times yield it's also the same but the difference in moment has to do with the arrangement unequal arrangement of these reinforcing bars around the perimeter each of these different symbols represents a condition in the the extreme layer of steel so I'm talking about the M the y-axis now so this point represents where the steel strain is zero in this level I'm highlighted highlighting with a pointer this point here represents when the steel strain is at 50% of yield at failure of the cross-section this is the balance failure point and these two other points are in the transition zone where the steel strain is twice that at yield and the steel strain is 0.005 entering into the tension controlled region so depending on whether the section is at a point that is compression or tension controlled you might recall that the phi factor in the ACI code the strength reduction factor depends on whether the section is compression tension or in transition so this solid line represents a tight column where points that are in the compression controlled region receive a phi factor of 0.65 points that are in the tension controlled region we're talking about phi factors of 0.9 and we linearly interpolate between those two extremes right between compression controlled and tension controlled so we have to reduce all the nominal values and I'm going to go back one slide all of these are nominal values we have to reduce them appropriately by the correct strength reduction factor and this is the spreadsheet I was mentioning this is just bending about the y-axis and the example is that the extreme tension fiber this the steel in the extreme tension fiber is at a strain equal to zero so that corresponds to this strain diagram so this is an arbitrary point so what I'm doing is just the calculation for one point and I'm trying to find well what axial force can this column take and what at what eccentricity which similarly stated it would be how much axial force and moment can this column take when the strain condition is as shown by this triangle so again epsilon s1 at the extreme layer is equal to zero as I arbitrarily chose and then failure crushing failure occurs at 0.003 so with those two points this one and that I can construct the strain diagram from the strain diagram I can compute internal forces stresses and forces for the steel bars in each layer we can also determine the depth of the equivalent stress block which is beta 1 times C. C in this particular case would be 18 inches minus the cover to layer 1 of steel and so this is the force diagram internal force diagram and all the calculations are shown here in spreadsheet format so you see this value is zero C is the neutral axis depth from similar triangles 15.5 right there A is beta 1 times C the areas of steel are areas that are plugged into the spreadsheets I had up to four layers calculated in this spreadsheet the depths are from for layer 1 2 & 3 are listed here and then these are the strains that come out from the similar triangles in this figure internal force values I just listed them as F 1 2 3 & 4 their values are shown here value F for F 1 would be for steel number what steel layers 1 2 & 3 are F 1 2 3 & 4 which we don't have a layer 4 here the distance to the moment axis which would be Y in this case is listed in this column and the compression force a resultant is calculated as shown here which is 0.85 F prime C times A and times the width of the section which is 18 inches so summing up all these values here gives me nine hundred and eighty five point seven kips that's the nominal strength multiplying this value times the distance to the reference axis Y gives me the moment so I have a total moment of twenty six hundred and ninety two kip inch if I divide M over P that gives me the eccentricity of 2.7 and for this particular point the phi factor is 0.65 so these are 641 kips roughly and 146 kip feet for the design values for axial force and movement so that's just one point so that's why it's important to do it in a spreadsheet because then you can change the condition the strain condition for the extreme layer of steel or set equivalently we can set an arbitrary location for the neutral axis depth and then construct these strain diagrams and then we can take as many points as we have patience and we start conducting you know building this interaction diagram from the concentric axial compression the point I just calculated these are the values from the previous slide and then other points and then we multiply them by the phi factor so to reduce them from nominal to design and this is how our design diagram would look like for this particular column cross-section all right so there are other other programs as well PCA SP column pardon me calculates interaction diagrams for reinforced concrete columns and you might want to construct your own Excel sheet or spreadsheet to construct interaction diagrams for pre-stress columns as well so this shows the effect of a bending about each of these axes these are the design diagrams for bending about the x-axis again in blue bending about the y-axis again in green pardon me and notice that the nice smooth shape that used to be the nominal now is not so smooth because of the change in the phi factor when you transition from compression controlled regions to tension control or transition region so that's why we have this little nose in the interaction diagram the other limit is that this is the maximum axial force that the ACI code allows you to apply to a cross-section it represents 80% of the design concentric axial compression phi times PN0 if your column is tied or 0.85 phi PN0 if your column is the transverse reinforcement is a spiral reinforcement instead if instead of constructing the interaction diagram you want to use design aids there are several design aids that are in the PCI design handbook and I'm just showing you one here for a reinforced concrete column so what's important to see is well what the shape of the column is and you see that this is this these series of design aids are for a 16 by 16 inch column with F prime C of 5000 psi mild reinforcement I believe that's part of the design aid figure and these three different curves show you when the section is reinforced with eight number nines four number tens or four number eights similar for this other column which is a little bit larger so you have interaction diagrams that have been built constructed using these properties when you have prestressed concrete columns what you have to account for is the effect of prestressing in the strain in the prestressing strand as I mentioned earlier so the interaction diagram curve looks different from from a reinforced concrete column interaction diagram so these are some design aids they are constructed here using four half-inch strands on an 18 by 18 cross section with different concrete strengths you see that the effect of concrete strength also has an effect of reducing the capacity of the section recalling that this is a failure envelope of the section and so you can use those for design right so any point that would plot inside an interaction diagram would be a safe point and any point outside of it would be unsafe so let's talk about some some limits reinforcing limits launch tool reinforcement ACI tells you that for columns with an average pre stress of less than 225 psi the reinforcement ratio has to be within this bound 1% of a gross and 8% of a gross so these are the limits the lens to reinforcement ratio ACI also tells you that transfers reinforcement must be provided using tie spirals or hoops and requirements for those are listed in these sections transfers reinforcement must also be provided for shear strength if needed and splices have to be accommodated as indicated by 1075 in the ACI code so I wanted to show you a couple of adequate transverse tie arrangements there might there might be others you have a longitudinal pattern reinforcing pattern you want to restrain those longitudinal bars with the corner of a tie or a hoop so this is a possible arrangement of ties the spacing of a ties and hoops cannot exceed these limits and a minimum spacing of 1.33 times the maximum aggregate size and the minimum tie sizes are number 3 bars when you're in closing number 10 bars or smaller and number 4 bars for number 11 or larger so notice that you do not need a cross tie here as long as there's an intermediate bar that's close to a restrained bar in a corner of a hoop or tie that is closer than 6 inches this is these are requirements from 25 7 2 in the ACI code so this is very briefly and I recognize very quickly summarizes interaction diagrams because I wanted to show you some calculations so we're going to go back to our design example and the column of interest is on column line a2 so notice that this column has a tributary area roughly where the pointer is moving here to mid span and halfway to the adjacent column line a2.3 so what I'm going to do in the next few slides is first calculate loads on this column and then go into the interaction diagrams to see if the column is satisfactory in elevation this is a footing this is again the same example building we have two stories it's a two-story column there's a little bit of an extension at the top I have some details this is an elevation along grid line a and these are the different elevations zero feet I'm taking at the at the ground level and then at level 2 this is the 10 foot height level 3 is that 20 21 foot height and a sketch representing this column along column line a shows you the relevant dimensions this detail here where the piece of use are acting the loads that are coming from the spandrel beam which in turn supports the floor double Ts are shown here so it's that the spandrels are embedded into this into this setback here in the column so that's this is this is a protrusion that extends outside of the column in our case so the axial force is got an eccentricity of 16 inches roughly from the column the column is going to be 24 inches by 24 inches in cross-section it's a 2 foot by 2 foot inch column 2 foot by 2 foot column partly so the first thing is to determine floor loads so I this is from module 1 where we calculated loads on the spandrel so I won't repeat it here these are all coming from the double Ts that are supported on the spandrel so these are let's say the stem loads and we the first thing I did is I said well these are the factored loads coming from each stem of a double T then we need to count the number of stems that that column is responsible for carrying so before doing that I did the same thing for between lines 2 and 2.3 which we hadn't done before in module 1 so again self weight of the the double T 3 inch topping that gives me the total dead load live load on the floor of the garage is 40 pounds per square foot so the total factored load per stem is 28 kips per stem between 2 and 2.3 and the values in this previous slide for between 2 and 1 so now what we do is well I counted the number of stems that that between each of the column lines that were tributary to this center column so if I go back and I won't do this in too much detail I will let you read through the slides in and see it so if you see here there's 1 2 3 4 5 6 7 about 8 stems up here and 1 2 3 4 so I tried to estimate how many of these double Ts would be carried by this column assuming symmetry of loading and I came up with these values so in terms of dead load from the floor loads I calculated six stems that would correspond to lines the total between lines 1 2 and 2 and 2.3 so 4 in one direction and 2 in the other apartment 6 above it and then 2 stems between 2 and 2.3 and then these 3 stems I I'm blanking out on what I did here but it's it has to do with oh I see it has to do with the distribution of if you remember correctly if you remember I had forgotten about this we have two different widths of double Ts that are between lines 1 and 2 one of them is a 12 12 foot wide double T the other one's a 10 foot wide double T so that's why these values of dead load and live load change depending on what double T we're dealing with and that's why so I counted the total number of stems of the 12 foot wide double T the total number of stems of the 10 foot wide double T and then I divided by by by 2 I mean this is this is this is the 10 foot wide double T the 12 foot wide double T then I divided by 2 meaning that between the lines 1 and 2 half of the load goes to the corner column half of the load goes to goes to the column we're designing and then I added to that the the 3 stems that are between lines 2 and 2.3 that's how I did the calculations similarly we did for I did for live load I also considered spandrel self-weight it's all dead load that's 34 tips roughly for at each floor level for the column in question and then we need to add the column self-weight so notice that all these loads floor loads and spandrel self-weight are loads that are acting eccentrically with respect to the column centerline and column self self-weight is acting concentrically and you can follow the calculations hopefully one thing I wanted to show you is design aid in the PCI design handbook you can either you know analyze this column and model it as a as a long column that has lateral support at each floor level or you can use one of these design aids as long as you're building a story heights are equal so this is a two-story column where equal story heights where this particular design aid 516.4 has this table that that gives you you know the number of stories in our case it's two and then it it shows you how to calculate these coefficients here represent the value that you would use to multiply times the eccentrically applied force so the actual moment at each floor level times these values case sub n to get the moment diagram so in other words what this is doing is the equivalent of analyzing this system using a structural analysis package under the application of a unit moment at the top of this column so if you analyze this using any structural analysis package apply a unit moment here and this column is restrained at the floor level levels and fixed at the bottom these are the reactions you would get and this is the moment diagram that you would get and if you see the coefficients here in the moment diagram 1.29 and 0.14 are shown here 1.29 and 0.14 where D is the point of application of the moment plus 1 C and B are essentially at the same place except they're just because of a sign convention they they have different signs the 0.29 here that's a moment and negative 0.14 is shown here at the bottom so these are moments induced by a unit moment applied at the top at the different levels of this column this is these are moments that are caused by the restraint of a floor system the first series of values here are values of restraint so you can see here that say for example the 1.29 1.72 and 0.43 come from this row which is assuming the column is fixed at the base you have 1.72 at them in the at level sorry I'm reading the wrong row 1.29 at the top is the restraint force negative 1.72 is in the middle level and 0.43 positive is at the base so that's what these values represent so we could use those to do the calculations by by hand I realize it might be more involved it might be easier to just use a structural analysis program the key is that regardless you have to separate what loads are acting eccentrically what loads are acting concentrically and in this slide I've separated the eccentric loads in floor number three and two and these are shown here on the right eccentric loads are coming from the floor system primarily right including the spandrel beam concentric forces are self weight of the column and these are eccentric forces at level three and two concentric forces at level three and two so adding all the total forces 237 at level 3.6 plus 5.76 gives me the total axial force the total force in level 2 is this amount plus axial forces generated at level 2 including self weight so all of these is the axial force at level 2 and notice that the moment in level 3 is simply just the moment induced by only the eccentric portion of these axial forces times the eccentricity of 16 inches and the same way the same observation for moments in level 2 so if we wanted to use that design aid in the PCI design handbook we could then say well use the the moments that we just determined in the previous slide of 316 for level 3 and 318 for level 2 and use the coefficients say for the two number of stories for this column we have m3 and m2 and the factors that we would end up using are when the axial force we have to recognize that we have to use factors when P is acting at each of the individual levels or we could use the addition of all of both of them I illustrate the calculations here separately so that you get a feel for what you're doing but you could also use this row which represents the addition of these two coefficients remember that the coefficients are coefficients representing values of moment for a unit moment applied at the top or applied at the second level in this particular second row so let's see let's just look at the first row and then I'll skip to the end so we have a moment in level 3 of 316.2 If the base of the column is pinned, we use the point B in this column would be here in the second level. So when M3 is applied at the top, the second level moment is 0.25. So this is, we're calculating moments in level 2, by the way. So 0.25 is here. Pardon me. And when the moment is applied in level 2, we would use 0.5 here. All right. If it is fixed, we would use this row in the fixed condition. And we have to make some assumption of partial fixity. So in this case, I use 60% fixed because there's no such thing as a full fix or a free column at the base. So we would have to interpolate. That's the interpolation that results. This would be the moment in level number 2, given the application of moment at level 3 and level 2. Similarly, for moment at level 3, we would have to consider the moment applied at level 3 and level 2. These would be the design moments. And finally, we have the final conditions of axial force at level 3 and moment at level 3, axial force at level 2, and moment at level 2. And using these two values, the combination of axial force and moment, we have to plot them in an interaction diagram for this particular column to see if they are appropriate. So we have a 24 inch by 24 inch column. Thankfully, there's one of these columns, design aids, in the PCI design handbook. So we plot that. It's a 5,000 psi concrete column. So we plot the two combinations of axial force and moment shown here. So we could essentially reinforce this column with four number 11s, which is the innermost diagram here. It would be these two points plot inside of that interaction diagram, meaning that the column is safe if we reinforce it with four number 11s. If I use the spreadsheet I was telling you earlier, I was showing you earlier, I have two reinforcing patterns. The one shown here with four number 11s is up at the top. That's the green line here. A more, slightly more efficient design, perhaps more difficult to build. Eight number 8s, it has a little bit less reinforcement, would give me that interaction diagram in blue. I plot again the two combinations of axial force and moment, showing that either one of these two designs work. And finally, the column details, just by looking at ties and so on, if we choose the four number 11 arrangement, we would only need a perimeter hoop. It would have to be a number 4 hoop, because we are using number 11 longitudinal reinforcement. And the tie spacing would be the closest spacing from these three conditions. So 22 inches would govern. All right, so that's the very quick and brief review of interaction diagrams. And I wanted to now go into slenderness effects. You know, what is slenderness in columns? And the first thing I wanted to ask is, you know, this is for the second poll of tonight, I wanted to ask what procedure do you normally use to account for slenderness effects in columns? So I'd like to ask Carrie to start the poll. Okay, the poll has been launched. Please answer on the screen. I will be leaving the poll open for a few more seconds. Please vote now. If you have not voted, please vote now. I'm going to be closing the poll. Okay, Sergio, I closed the poll. Forty-three percent said I analyze structures using nonlinear geometry. Thirty-six, I use the moment magnifier method. And 21 percent, I have never considered slenderness effects. Okay, thank you, Sherry. You're welcome. So what we will do is, those of you that are analyzing structures using nonlinear geometry, p-delta effects, that's fine because... So these are the approaches that the ACI column endorses. So the first thing to ask ourselves is, when does one need to consider slenderness effects in compression members or in columns? And the ACI code says that whenever you have a significant reduction in axial load capacity, about five percent or more, you have to consider slenderness effects. The ACI code especially says, or particularly says, you have to account for reduction in axial load capacity due to moments resulting from lateral deformation of the column. And I'll go into that in the next slide. The ACI code endorses two approaches. One is using an elastic second-order analysis. That's often referred to as a p-delta analysis. And the second approach is to do an elastic first-order analysis without p-delta and use moment magnification. Those two are acceptable approaches. The second-order p-delta analysis involves analyzing the structure or calculating its equilibrium with the loads acting in its deformed configuration. It is the most accurate method to do it. However, you have to use appropriate stiffness considerations and make sure that you understand how each individual program is doing a p-delta analysis. Because sometimes you may get results that are non-conservative if you don't know what you're doing. The other thing is that when you consider slenderness effects using p-delta analysis for non-sway columns, so those columns that are not part of the lateral system of building, you have to make sure to add nodes, intermediate nodes, between the ends of the columns because otherwise the slenderness effects for a column that doesn't sway laterally are not captured in general by structural analysis programs. P-delta analysis that are in structural analysis programs are typically for columns where there's lateral displacement between the top and bottom of the column, so the column that sways. The second method, the moment magnifier method, is what I will be discussing here in more detail. And it consists of doing a structural analysis, first-order analysis, no problem, with the structure in its undeformed position and then using those moments and amplifying them using the moment magnifiers or delta values or factors that are in the ACI code. So what does it mean to amplify moments? So consider this column that has an axial force acting at an eccentricity from its axis. So you know that in its undeformed shape, this is just for reference purposes, this is how the column is represented here, it has a little eccentricity of axial force. So if you consider the undeformed configuration, you would simply apply, well, axial force goes through this line and it doesn't produce additional moments. The only moment is the moment that acts at the end of the column, that's the P times E moment, that's referred to as the first-order moment diagram. It's constant moment throughout the length of the column. However, if you look now at the deformed shape of the column, after application of this eccentricity or axial force at eccentricity E, there's an additional component to moment because if you take a free body diagram about the deformed configuration of the column, you have this added moment which is equal to P times V0, V0 being the lateral deflection of the column at that level. So this is the second-order diagram, moment diagram, that only appears if you consider the deformed configuration of the column. So these here are deflections, lateral deflections of the column. Notice that the total moment is PV plus PE at any point along the column, or the maximum moment is PV0 plus P times E. So this is a first-order moment, P times E, and PV0 or P deltas are the second-order moments. And what effect does that have? So recall that from our discussion of interaction diagrams, if you have a constant eccentricity and it's not changing, you're loading along this line until you reach the interaction diagram. When you have an eccentricity that is changing because of the lateral deflection of the column, you're loading non-linearly because every time you increase P, it increased the lateral deformation of the column. So you have a compounding effect of P and eccentricity that has this additional P times V moment. This is the P delta moment. And eventually you're going to cross the interaction diagram at a lower point than you would have if there were no additional eccentricity. So you have a point A, or the moment magnification. There are two ways to consider this problem, either by magnifying the moments that you account for in design, or by reducing the axial force that the column can carry. Those are the two approaches that are the two effects that second-order effects have on columns. So that is, this is really called, it's when these lateral deflections that are significant that we need to account for slenderness effects. So when you have unequal end moments, not the uniform end moments, that means that the axial force is acting at different eccentricities, you simply have a first-order diagram, first-order moment diagram, with unequal end moments. But you still have this lateral deflection that occurs, and that generates this lateral deflection that generates this second-order diagram. And notice that in this particular case, the maximum moment doesn't necessarily occur at the center of the column because of this unequal distribution of moments. So first-order diagram is here, second-order diagram. M2 will always be denoted the maximum end moment, M1 is the minimum end moment, and this is a column that is not subjected to side-sway. So in other words, it doesn't displace laterally. So in summary, the moment magnifier approach consists of using the first-order moment, the maximum first-order moment, M2, and then amplifying it by some factors such that we get the design moment, MC. So this amplifying factor, amplification factor, is delta, and we use that to amplify the larger end moment. L sub u, by the way, is the unsupported length of the column. Okay, so for the moment magnifier approach, the ACI code uses the following definitions. It defines non-sway frames as those frames where lateral displacement is negligible between two floors, two consecutive floors is negligible, or whereas, you know, and it's described as a stability index Q less than 5%. I'll talk about the stability index later. Sway frames are those where the displacement between two consecutive stories cannot be neglected, right? It's appreciable. So in order to fall into these, whether a frame is sway or non-sway, you have to check a column on a story-by-story basis. So you cannot just say, okay, this entire story is, pardon me, this whole building is sway or non-sway. You have to check a story may turn out to be a sway story if there's appreciable lateral displacement within that story, even though the rest of your building might be non-sway. Okay, so you have to check the stability index for each story in the building. And the first step in calculating slenderness effects is to understand what, how this lateral displacement takes place, even with the application of an axial force. You might remember, those of you that have taken strength materials a long time ago, such a thing as an Euler buckling load. I mean, I often make the example is, take a ruler, a slender ruler, and press it down, and you will see that eventually you'll reach a force where it starts buckling sideways. And that's the typical example of the Euler buckling load. I won't burden you with the math. I leave it here for those of you that want to look into it, to remind you of it. But what's important is that when you reach that level of axial force, it's called the critical buckling load, you know, when you reach it. And it can be calculated by using pi squared EI. E is the modulus of elasticity of this column. I is its moment of inertia, divided by the unrestrained length squared, L sub u length. This particular solution is, it's only applicable to the case, this particular case of a simply supported column as shown here with L sub u as denoted here. So in general, when we have, so this is for a pinned-pinned column, those are called the boundary conditions. So this solution here is sensitive to the boundary conditions, whether it's pinned-pinned or fixed-pinned. So in general, instead of using this Euler buckling load or critical buckling load, the codes generally express this unrestrained length as the effective unrestrained length by multiplying it by k times L u. So it's this k L u squared term in the denominator that's important. So what does that effective length correspond to? So it's for a column that has pinned-pinned ends, k would be equal to 1. So L sub u is simply 1 times L sub u, or k L u is 1 times L u, so that's L sub u. Let's say you have a column that's fixed-fixed. The points of inflection or contra flexure resemble the shape of this simply supported column. So the distance between those inflection points is often, you know, physically understood as the effective length of the column. So that's called k L u, and in the particular case of a fixed-fixed column, this distance is equal to one half of the clear unrestrained length of the column. So k for a fixed-fixed column is 0.5, k for a fixed pinned column is 0.7, and k for a condition that is not fully fixed or pinned, so elastically restrained as such as by beams, is some fraction between all these limits of 0.5 and 1. So this k factor is determined using alignment charts that you may have used for, if you've ever conducted steel design. So this is, by the way, for columns that don't have lateral displacement relative to bottom to top. For the cases of columns that do have lateral displacement, k exceeds 1. So this is the best example is, say, if you have a fixed pinned column that can displace laterally, imagine the shape that it would take to resemble a simply supported column. So the k in this particular case would have to exceed L sub u. In this case, it is 2 times L sub u. So the key feature to remember here is that for columns that are not swayed, that are not restrained against side sway, you have a k factor exceeding 1. And for columns that are restrained against side sway, you have k factors less than 1. So this is the moment magnifier equations in a nutshell. So the kLU, so this is the slenderness effect of the slenderness ratio of a column, kLU over R, I will define R in a second, has to be less than this quantity here, or less than 40. And if that's the case, then the column is not considered slender. That is for non-sway frames. So these are columns that are restrained top and bottom. If you do not have to consider slenderness effects because the slenderness ratio is less than this quantity, then you're done. You just design the column as if it were a short column and you use interaction diagrams. If you don't satisfy this condition, then you have to amplify the maximum end moment by the magnifier delta, which is calculated using this equation, where P sub c here is that Euler buckling load in which EI is then the product of E times I is used as EI effective for to use a reduced stiffness for concrete. We'll talk about the details here in a second. So these are the equations that are relevant for now. So this is the amplification factor, and P sub c is the equivalent of the Euler buckling load. C sub m is a factor that I'll explain later. So it's important that these equations rely on m1 over m2, which is the ratio of minimum moment to maximum moment. And there are two conditions when the column is subjected to moments that are of opposite sign, as shown here, are single curvature, but one of them is clockwise, the other one's counterclockwise. In this case, m1 over m2 is to be taken negatively, with a negative sign. In this case, it's to be taken with a positive sign. If you're using a moment, pardon me, an ACI code that is older than ACI 318.14, the definition of single or double curvature sign for the ratio of m1 over m2 is exactly the opposite of what's shown here in this slide, just for you to be aware. If you're using ACI 318.14 or later, this slide is correct. The moment amplification factor delta is given by the equation I showed you before. C sub m is a factor that is used to converge unequal end moments to equal end moments. The actual moment amplifier is shown in brackets here. You know, it's just a little detail here of the amplification factor. What does that Cm factor represent? So, this is the general condition of unequal end moments, m2 and m1, where I indicated earlier that the maximum moment may not occur in the center. For the case of equal end moments, the maximum moment, including second-order effects, does occur in the center. So, C sub m is a factor that's used to convert unequal end moment condition to equal end moments, for which we know what the amplification factor delta is. So, this C sub m is simply a conversion. And you'll note, though, that the maximum moment does not occur at the same place in these two conditions, the actual condition and the equivalent condition. But it doesn't matter, because we're designing for a maximum amplified moment and an axial force in the column. This figure is taken really from the first set of codes that used slenderness effects using the moment magnifier were a design of steel columns. So, the C sub m factor approximately is calculated using this equation. The real values are shown by these series of slide or solid lines that are a function of the axial force acting on the column divided by the critical buckling load. This equation, which is really a line with a negative slope, is shown here, and it cannot go under 0.4. So, it's an approximation to actual solutions for C sub m. So, again, the ratio of m1 and m2 appears here. And in order to calculate the buckling load, we need to know what L sub u is. So, these are three different conditions. L sub u is the clear unsupported length of the column. And we need to use EI effective to account for cracking of concrete. The K factor is, as I alluded to earlier, is determined using alignment charts. Okay, so how do we determine those alignment charts? We have to, or the K factor, these are two alignment charts for columns that don't sway laterally or those that do sway laterally. This is a reference taken from ACI 318, where you calculate the ratio of EI over L of columns divided by EI over L of beams. If beams are made continuous with the column, this is for reinforced concrete design. So, if there's restraint from beams, you know, these elastic restraints, you compute the ratio of EI over L of columns to EI over L of beams. You calculate this psi factor for the bottom of the column and the top of the column. You graph that here for the bottom and top. You join those two points with a line, and that would be your effective length factor of about 0.76 in this case. And for the EI effective, for these calculations for I, the ACI code tells you, well, if you're dealing with a column, account for some cracking. Use 70% of I gross, and for beams, use 35% of I gross. So, this table is in ACI 318-14. It gives you a moment of inertia that one could use to calculate these restraint conditions for the alignment charts. By the way, if you're doing a p-delta analysis, you should also use moments of inertia that are similar or, you know, somewhat account for cracking of beams and columns. And the ACI code also gives you several equations for EI effective to account for cracking and the presence of steel. The more detailed equation is less conservative than a simpler equation. This simpler equation where preliminary design, work for preliminary design is fine because it doesn't account for the EI effect of longitudinal steel. This one does. And the denominator here accounts for sustained loads. Beta DNS is the sustained load factor, which is the ratio of factored sustained loads divided by factored total loads. So, I'll let you look at the ACI code for the definitions of these factors, but that's what it is. This is a relatively new equation in ACI. It also allows you to use EI effective by using ECI, where I is computed at different table 66311B instead of the one I just showed you earlier. And you can use this expression to compute EI effective in calculation of the buckling load. So, this shows you, this slide shows you the degree of conservatism in the two EI effective equations. This is the first one that was shown in, the second one, pardon me, that was shown in the previous slide, which is shown here. It is much more conservative because the values for EI vary between 2 and 5, so roughly, and the ideal fit or the accurate EI value would be at 1. So, it is really conservative. So, it is mostly useful for preliminary analysis. And very briefly, because we're running out of time, I'm just going to cover a couple of slides on slenderness effects for sway frames. The slenderness effect, the slenderness ratio cannot exceed 22. And in this case, you have to separate, when you're dealing with sway frames, you have to separate loads that do not produce sidesway, and that's shown here. So, you have to design the column for the both end moments. Moments that do not produce sidesway are not amplified, and moments that are produced sidesway are amplified by these delta S factors on the second term of this equation. So, you have to separate, essentially, gravity moments from lateral load moments in the sway case. And you have a set of similar equations to calculate the amplification factor, where the stability index Q is defined here. I'll show you in the next slide what this means. Or, you can use a similar expression as before for the non-sway case, except that you have summation of axial forces divided by summation of critical loads. Or, finally, use second-order analysis, as some of you already do. The summation factor has to do, and also here in the stability index factor, has to do with the fact that for sway frames, a column cannot sway laterally if the entire floor is not swaying laterally the same amount. So, the lateral sway is denoted here as delta 0. So, you'll notice that all columns in a floor sway the same amount. That's why the summation quantity is here. So, that's delta times summation of P sub u. It gives you a destabilizing moment, and the shears that are generated at the top and bottom times the distance between them gives you a stabilizing moment that equilibrates this system under lateral sway. And that's how the stability index is defined. Okay, so these two slides, this is one of two and two of two, is how one uses the moment magnifier approach. I wanted to show you, you know, in a series of bullets what to do, right? First, we have to separate loads that don't cause side sway from those that do. That's the first step. We determine in moments, right? M1 and M2, axial load, lateral displacement, account for cracking of structural members. We calculate the ratio of stiffnesses of beams and columns at the end of the members to determine the effective length factor. Use the alignment charts to read the effective length factor. Establish or calculate the slenderness ratio and determine whether slenderness effects need to be considered. If it needs to be considered, we have to calculate moment magnifier for non-sway and sway frames, depending on what case we're dealing with. And we multiply the largest end moment, M2, from the first-order analysis using the corresponding magnifier for braced frames and both moments for unbraced cases. And then we design the moment using the axial forces determined from the analysis and the factored magnified moments that are used from the slenderness case. So going back to the design example, I looked at our building and the columns did not qualify as slender, so I went to the middle wall. So this is only to just illustrate slender column calculations, but I'm designing, I'm just showing it with respect to this center wall right here, because it was the only slender element that I found in this brief building. So the first thing to do is look at the geometry. Notice that the solid portion of this wall is like a column. This is the cross-section. It's at 8 inch by 28 inch pier, and loads are acting directly on top of it, or at least that's what I assumed. So this could buckle, could be considered a little column. The side elevation, this is a three-story column with axial forces coming down from the floor system. Notice the slight difference in elevation for the brackets here. And we recognize also that that might be unequal loading, right, on either side of this wall. So the first thing would be to calculate what the dead loads and live loads are. So I calculated the dead loads for one of these brackets here for each stem. So notice that each of these is one stem of a double T that's being supported, so that's the axial force that's acting. So I went ahead and calculated the axial force due to dead load, and then in order to maximize moment, I considered live loads being applied on one side of the wall and not the other. I just paused because I believe I told you that load patterning should not be done. In a garage, you can do load patterning. What you can't is reduce, you can't reduce live loads. I don't know if I said that incorrectly, but it just dawned on me. So live load we're patterning, we're saying, I'm thinking, well, let's say that live loads are only acting on one side of this building and not the other to maximize moment in addition to axial force. So we did dead loads and live loads, and I'm going to assume 60% fixity at the base as I did before in the examples that I went through earlier. So again, using the design aids that are one of the design aids that's in the PCI design handbook, now we have a three-story pseudo column. It's really a wall, right? It's a wall pier. And so this would be for pinned and fixed, so we have to interpolate between the two values. And what I'm using here is the summation of all the moments. The effect of moments acting at the three levels here is given by the summation row here in this table. So what does this mean? It means that at point B in this column, which would be right here or in this wall, I should say, the total moment produced by the action of potentially applied moments being produced at level B, level D or E, and F is represented by this 0.67 coefficient K sub m. Okay? So the 0.73 then represents, say for example, that would be the moment at point C would be that one right there, just above the bracket. And similarly at point D and point E, when moments are acting at the three levels of this particular structure, right? For levels 4, 3, and 2. It will be 4, 3, and 2 right here. It's a slight difference in the way I labeled the floors from what the table says. So again, these would be the coefficients that would be of interest. We need to calculate P times E, the eccentric axial load. So let's say we want to look at the moment at point D here, at level 2, at point D. So we consider the axial force acting at level 2 through 4, the three levels. From the previous slide, we get a K sub m of 0.92. We estimate the eccentricity as four inches, that's half of the wall thickness. Three quarters of an inch is the gap between the end of the double T and the face of the wall. And two and a half inches is one half, this arrow should point to the two and a half inches here, one half of the pad length. That gives me an eccentricity of 7.25 inches. The axial force that's coming from a single stem of a double T to the wall is 10.10 kips. At that eccentricity and the K sub m factor gives me a moment for design of 67.4 kip inch. I could do the same type of calculation for point D. So the axial force at point D is 91.9 kips. This would be for level 2 at point D. I could do the same calculations for point 1, which is point, or level 1, sorry, point B in the sketch for the design aid from the PCI design handbook. So this would be the first order moments and this is the axial force. So this is essentially my structural analysis equivalent. So I need to check whether I need to amplify this moment because of slenderness effects. By the way, your slides have a slight error here. It listed 53 something. It should be 52. So we first check, well, do we need to consider slenderness? So let's calculate the slenderness ratio for this wall pier. R, the radius of gyration, is defined as a square root of I over A for the wall pier. 28 inches wide, 8 inches thick and divided by 12 is the I and the area is 28 by 8. 2.31 is R. So assuming a K of 1, which is conservative for a non-sway condition. L sub U is the unrestrained, so the floor-to-floor length minus the depth of the double T. So I'm considering that the double T may restrain the wall top and bottom. So I'm subtracting that in the overall height and then I calculate KLU over R, which is 43.2. It exceeds 40, so I have to consider slenderness effects. Also that's given, so if I assumed just a point that the bending that would be occurring here given the moments would be single curvature bending. So it would be negative for single curvature bending. So my KLU over R would have to be 24.8 or less for to neglect slenderness effects. Since I have 43.2, I have to consider that. So the first step is to calculate the critical buckling load. I squared EI effective over KLU. E for concrete is 5,000 psi square root, 57,000 is for a normal strength concrete. That's E for concrete. I gross, I'm using the simplified equation for EI effective. I gross is 1,195 for that wall pier. 0.4 is just to account for the reduction in stiffness in general. Remember that this is the conservative equation. And P delta NS or beta DNS, pardon me, is the sustained load factor, is factored sustained loads divided by ultimate factored loads. That would be 71.7 for 1.2 dead and 91.9 for 1.2 dead plus 1.6 live. That's an amplification long-term effects factor. So that EI effective is over 1 million. Plugging that into the critical buckling load equation and assuming a K of 1 conservatively again for a non-sway column with the clear span of 99.8 inches gives me a thousand kips for P sub C. And then using to calculate the magnification factor, I have to calculate CM first using the ratio of M1 over M2. That's the ratio again. Negative for single curvature bending. So this negative times negative would become positive. That's 0.91 now. So 0.91 divided by the factored load here and divided by 0.75 times P sub C using the amplification factor equation gives me a slight amplification of 3%, which I use then to amplify the maximum moment that I determined from a first-order analysis using the PCI design handbook. So that gives me 69.2 kip inch that in combination with the axial force would I would then have to use to design to plot the combination of this amplified moment plus the axial force in a corresponding interaction diagram for an 8 inch by 28 inch pier to see if the column is acceptable. I didn't do that because because of time but that's what one would have to do use this amplified moment in our design instead of the first-order moment of 67.4. Finally there's one last check by the way that the ACI code asks you to determine is to consider the minimum eccentricity of 0.6 times plus 0.03 h for a cross-section. So 0.6 plus 0.03 times 8 would give me a minimum eccentricity that is larger than what I have done before so than what I've used before. So the minimum design moment in this particular case would have to be 77 kip inch instead of 69.2 kip inch by using the minimum eccentricity required by code. So this would be my design values that I would need to use to determine the reinforcement within the wall pier for that particular example. So as I said there was a lot of material to cover in this session and we barely made it so hopefully you'll I know it's a lot of material hopefully you'll get a chance to review on your own after this session and I'll be happy to answer any questions you might have right now in the last five minutes or that you sent to Sherry over the week. If you have any questions please type them in the chat box now. Well Sergio I don't see any questions as of yet. Oh I do I have one question here. Okay here's the question. The question said you said it's 52 instead of 53.5 but that's for point C and not point B. Let me go back to that slide. I was looking at the slides earlier today. It's 52 because what I would do is I would I would use the factor that I would use would be. Hold on let me just check here for a second. I believe I used. So in this slide you have the summation of for all the the effect of loading on all the points along the column and I believe I used this 0.73 or interpolating between 0.73 and 0.69 which leads to 0.71 if I remember correctly but I'm not finding that number here right now. Yes so it's point C right here is 52 and that's what I was trying to remember to calculate so point C would be here so we would have to use this 0.73 interpolate between 0.73 and 0.69 which roughly will give us a factor of 0.71 for K sub N. If you then do 0.71 for K sub N times 10.10 times 7.25 that leads to 52 kip inch I believe. So you can please check that calculation to make sure that I'm not so it's about 0.71 here. Question what is H in the minimum eccentricity equation? Okay so H in this case I didn't explain it clearly. H is the depth of a section so H in this particular case is the thickness of the wall. Remember that we're designing this wall pier so H is said is perpendicular is measured perpendicular to the axis of bending so if we have a an 8-inch thick wall since the loads are applied in the short direction of that wall the H would be considered as the thickness of the wall 8 inches. Oh somebody just replied says okay because it says point B instead of point C. So somebody found another oh yes I see what you're saying okay thank you. This is right thank you so the error is there's also another error this should say point C and not point B. Point C is is what this 52 kip inch is. Okay yeah so thank you yeah. Okay if you have any other questions we're about to end the session. Please send them in an email to me at snon.pci.org and I will get them to Dr. Brenier and we will get the answer to everybody during the week. Thank you so much for attending and thank you Mr. Professor Brenier. Thank you and I will see you all back next week at at 6 o'clock Central 7 o'clock Eastern. Thank you so much for your time.

axially loaded members

columns

compression

bending

interaction diagrams

slenderness effects

ACI code provisions

eccentricity

axial strength

design aids

calculating loads

moment magnifier approach