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Send collector calculated

hey everybody in this video i'm going to be showing you how to calculate the seismic force on a structures diaphragm this example problem falls under section 3 seismic forces for building structures of the board's test plan so let's get started first off what's a diaphragm what's a cord what's a collector well simply foot simply put a diaphragm is a horizontal element that takes seismic force generated in your structure and transfers that through in-plane shear to your lateral force resisting system whether that be a moment frame or shear walls or brace frames etc uh in this case for this problem we're presented with some diaphragm force we've got a flexible diaphragm that's transferring load to special reinforced concrete shear walls and it's doing so with the help of a couple collectors 60 foot in length on either side that i'm going to go ahead and highlight here okay so a diaphragm it can bend in plane and what that looks like is i'm going to go ahead and just draw an orange here when the load is generated in the diaphragm for a flexible diaphragm it's going to cause it to bend in plane just like so right kind of like a deep beam and what that happens you're going to expect just like a deep beam or just like a beam to have some compression in the top there right and then some tension there in the bottom and so typically you're going to want to be checking that perimeter element or designing a permanent element specifically to take that axial load okay and so this problem is asking us to find what that maximum force is and the force appears in the chord element so just like that top element you have in a truss that's seeing either compression on the top or tension on the bottom of that truss as the truss spans over some length you're going to have a cord element at the top and the bottom of your diaphragm if you will so this problem is asking us to find that maximum chord force along the north south edges so the the top here and the bottom here as i've drawn it out and then also to find the diaphragm shear force right so we're given the force f sub px in the north side direction and it's asking us to find the diaphragm shear force and drawn in red that diaphragm shear force is going to act along the edges here it's going to transfer that load to the lateral force resistance system as we said that is special reinforced concrete shear walls okay all right so we've covered two of the three that i mentioned i mentioned diaphragms chords and collectors okay we covered diaphragms we covered quartz and now we're going to talk about collectors so what is a collector a collector um it's exactly what it sounds like it is right it collects load from the diaphragm it brings it back to the lateral force resisting system and so you can see that we've got two collectors here on either side they're both going to be working to transfer load back to the lateral force resistant system and if you think about these collectors in terms of their boundary conditions you've got these concrete shear walls at the top and at the bottom of them you can think of them almost like restraints think of if you had a beam and an axial load running across that beam okay you're gonna have tension on one end and compression on the other end as that load is pulling on that beam and as you've got a restraint on one end and uh or a restraint on both ends i should say and then reactions at both ends right so we'll see in a second here how taking that shear force and that shear force being applied to the collector element is going to result in change in the load in that element going from a tension to a compression okay so uh we've got the diaphragm force of 10 kips and we're just going to jump into solving this problem here diaphragm force of 10 kips it's found typically specifically f sub px it's going to be found through the methodology labeled out in asce 716 section 12 10 on diaphragms chords and collectors right and so uh this problem is set up such that somebody would have already found that load f sub px and now we're going to take that and we're going to use it to calculate the maximum chord force okay so the way we find the maximum coordinates as i said is we're taking that in plane bending as we drew in the orange just then and we're using that and we're going to resolve that in plane bending until with some tension and compression so how do we quantify that in plane bending well we quantify bending through a moment of course so what is the maximum moment well we've got a point load here okay and how are we going to find the maximum moment on a beam with a given point load well it's just a commonly known moment equation a moment the maximum moment on a simply supported beam with a point load in the center it's going to be max is equal to pl over four right here l is a 180 and p is 10 kips right so i'm going to go ahead and calculate here m max is equal to 10 kips times 180 foot and that's going to be divided by four p l over four and so that load is going to result or that moment is going to result to be 450 kip foot okay so we've got the moment now how are we going to take that moment and convert it into a tension and a compression well it's just about the couple right same as you would do with a moment couple any moment couple you have you've got a moment about a certain point you can resolve it based off of that depth in this case the depth of the diaphragm is 160 feet it's 50 plus 60 plus 50. and you can go ahead and calculate your tension which is going to be on the bottom of your diaphragm your compression at the top that couple you're going to calculate it like so tension and compression is equal to 450 kip foot divided by 160 feet okay and so that's going to give us what that's going to give us 2.8 kips okay all right so we've got the maximum chord force in tension and in compression okay now there's one thing that we haven't done though yet okay that force it needs to be multiplied by some over strength factor as specified by section 1210 of code over strength factor being omega okay and specifically for special reinforced concrete shear walls you can look into the code and you're going to find that the over strength factor is going to be 2.5 okay so if we re if if we want to actually calculate what this is in terms of what we're going to be using for design because this is seismic design category d we're going to be required to multiply that by the over string factor so just get it back up here and i'm just going to go ahead and multiply that by 2.5 okay so 2.5 times uh what we had there which was a 2.8 and that's seven kips seven kips is our design force and just a reminder this is our over strength factor okay next up what's the diaphragm shear that's v sub d diaphragm shear as we said the diaphragm shear is what's going to be running along each side of the diaphragm the shear that runs along each side of the diaphragm transfers the load to the lateral force resistance system drawing it just once again there and so how do we calculate that well we know what our diaphragm forces is f px is equal to 10 kips right and that's running along both sides so if we divide that by two and we divide that by the length uh of the diaphragm or the in this case you could say the depth of the diaphragm if you owe 160 feet we're going to come up with the diaphragm shear okay so it's just v sub d is going to be equal to that 10 kips there and if you just go ahead and divide that 2 times 2 times 160 foot and remember the 2 is because it goes to either side where we have our lateral force resisting system elements the walls in this case so 10 divided by 2 and then divided by the length 116 and we're going to give that response in pounds per linear foot okay so that comes out to 31.25 plf now notice i did that conversion from kips to plf in in my head just there right so um next we're going to look at calculating the collector force right so as i said the collector force is what's going to it's going to come up it's a maximum force it's going to occur in these elements on the sides over here their collector elements are going to transfer the loads back to the shear walls okay all right what are we doing so uh if you try to think about what exactly is happening here the load gets transferred back to the shear walls the shear wells are really taking all this lateral load and so you have some shear resistance in the opposite direction and that's v sub wall if you will okay so what we can do here is if we go ahead and we calculate based off the load that's being taken uh by the shear walls and the difference from that to the diaphragm shear calling that the net force along this wall line we can visualize where exactly the maximum force in the collector is going to occur now just to make that clear here i'm going to draw in pink wall is going to be taking obviously at the end visualizing the force it's going to be zero force at the end here but visualizing the load along the wall we're going to come up here to a certain peak right because obviously v wall is going to be greater than v d so we're going to come up to a peak here and then that peak is going to quickly come back down to another peak on the either end on the uh on the other side over here as we consider the force v sub d and then the wall is going to kick back in we're going to get more resistance on that end and so we're going to have a tension on this end and a compression on this end okay and so we can quantify that by first finding what v wall is and then uh using that value to calculate using that value and the diaphragm shear value to calculate what exactly the maximum collector force is so just walking through that v wall is going to be equal to what the wall is going to be equal to that load f sub p 10 kips divided by two and then divided by a hundred foot of wall on either side so you see here we've got a hundred foot of wall on on either side of this okay so we've got 10 kips and we're going to divide that by 2 and then 100 foot and that's going to give us 50 plf okay so we know what the share in the wall is right and we know that 50 pounds per linear foot is resisted in the wall so like i said we're going to starting out from the end here 50 pounds per linear foot is resisted by the wall so drawing that force out we end up with a load uh with one sign if you will the sign is going to be obviously opposite to the v sub d because it's acting in the opposite direction and then we got no wall in the middle where the collector is so the sign flips to where we're going to have a compression in the collector and then that comes back down to zero where we have the wall resisting v sub d okay and so if we want to calc that this value over here the value of interest the tension or the compression in this situation since the walls are equal length 50 feet 50 feet they're going to be the same value equal but opposite okay we can quantify that by taking our 31.25 pounds per linear foot of shear along the diaphragm or 50 pounds per linear foot that are resisting resisted at the wall taking the difference of those two which ends up giving us 18.75 pounds per linear foot that's going to be the net force in this section over here the net force in that section over there and multiplying it by the length and that's going to give us that peak and just to show that to you here so that's going to be p collector and we're going to go 50 foot times v wall the wall is 50 pounds per linear foot minus v sub d which is diaphragm 31.25 pounds per linear foot okay and so that's going to give us what 93 937.5 pounds or 0.94 kips because we like to present uh load in terms of kips right um so there's one thing that we're missing and what's that again it's the omega force right so just going back to the same way we calculated the chord force we are missing the omega force from this collector calculation so backing up i'm going to erase that i'm going to multiply that by 2.5 make it clear that this is the omega and coming back down it's going to be that .94 and we're going to multiply that by that 2.5 and so we're we're looking at roughly about 2.3 kips okay and that's going to be the collector force right so just walking through that once again we have a diaphragm force the diaphragm force uh and along with the dimensions of the diaphragm in a flexible diaphragm allows us to calculate the maximum chord force okay it also allows us to calculate the diaphragm shear that's the shear that is along the diaphragm the shear that's transferred into the lateral force resisting system element in this case the special reinforced concrete shear walls and then we find the collector force again the collector is an element that collects that shear load it brings it back to the lateral force resisting system okay the lateral force resisting system in this situation is going to be the special reinforced concrete shear walls and so that's resisting all of the shear load on that side and using the uh the shear the resisting shear v sub wall and using the diaphragm shear v sub d we're able to calculate the net force along that wall line and then quantify what exactly the force is in the collector and we can see in this situation that the maximum force occurs at the start and end of the collector but the signs are opposite now just be aware that depending on what the length of wall is in this situation the signs are the values are equal but they are opposite but surely if you had a wall on one end that was 10 foot and on the other end was 50 foot of course you can imagine that on the end where it's 50 foot you're going to see a much larger more substantial amount of load as would make sense because obviously more load is going to go back to the stiffer element which is going to be the longer shear wall okay so this example is going to be on my website www.structural.wiki it's going to be there with the solution as i have it as i show it here and it's also going to be there as a blank problem and then you can fill in the blanks to kind of give you some practice there okay all right i want to thank everybody for watching these videos and feel free to check out my site and my email is on there feel free to shoot me a message or feel free to send me a message here on youtube and if you have any specific questions i'm happy to help thanks so much have a great