Transcript for:
Structural Analysis Key Concepts

hey and welcome to another adventure in the fe civil review so tonight we're going to get started and we are going to go ahead with a structural analysis so actually in the affe review there's a couple things going on here so what you can see here is actually this is a structural engineering section we're sort of staying on the structures topic so we've already done statics strength or mechanics and materials depending on what you call it but right now we're going to jump in and we're going to go ahead and we're just going to kind of keep going with structural engineering oh and the way that i see this is i kind of break it into two sections normally in a civil engineering program you're going to be taking like a structural analysis type of course and then you're going to take some design level courses on top of that as well so like concrete or steel design and some of you won't take concrete or steel design so we'll get to that next week but this week what we're going to do is we're going to jump in on some structural analysis problems so if you don't have the problems yet take a look at the link below and there's access to those problems with a pdf that you can follow along with and you can print those out or do them on the screen or in a notebook uh whatever you like but we will just get started here and hopefully this makes a little bit of sense because what we're going to be looking at is the statically determinate beams columns trusses and frames and a lot of this is kind of a carryover from a statics class so similar things that you'd be looking at in statics but it takes it a little bit further maybe the next step maybe it helps you to see it a little bit a little bit more advanced and we'll go into deflections and we'll do a couple different methods here i will look at the principle of virtual work kind of briefly for for trusses we'll look at some basic beams and and those sorts of things we'll get into columns columns is interesting because actually there's some column formulas in the mechanics and materials section of the review handbook but it shows up here in the structural engineering piece and even some indeterminate structures we'll take a look at we're not going to get into like super crazy like really long and complicated processes but we will take a look at some of the ones that are in the manual and go from there so without with that being said hey if you have comments um drop them in the box and we will just get started here so oh so went to a little too far here but all right the first question this looks an awful lot like a statics question that maybe we saw last time but what this reads is the axial force in member ec so what are we looking for we're looking for member ec so right here this this number here the axial force in that member due to the applied loads shown in the trust below is most nearly so to remember with trusses we're not dealing with self weight we're going to ignore the self weight and what we're going to do is we're just going to take a look at that one member so this truss is a little bit more complicated than some of the ones we've looked at whether you know some of the trusses will have parallel cords and what you can see here is these you know top cord and bottom cord they're not parallel so it it makes it a little bit more complicated you can't just do a sum of forces in a y direction or x direction to get a component here you really kind of have to use uh the sum of moments formula and i like this question for a number of reasons one because it really focuses you in on that some moments question but two if you think about it right and you start to understand trust is a little bit better it it's going to help you with this process of solving them so with them this is going to use the method of sections and to do the method of sections what i like to do is i like to kind of draw a section around a part of it and the reason i'm drawing the right side is just because it looks simpler to me okay so the right side looks a little bit simpler simpler i kind of like this little this little truss in here right uh and you know it just looks a little bit a little bit simpler so what i'm going to do is i'm going to first just draw a free body diagram of that and then we're going to take a look at it again so i'm just going to copy this down here to draw my free body diagram all i'm doing is kind of redrawing this thing uh and i'm not probably drawing it perfectly here but just redraw on that section and what i like to do is just take that the yellow triangle first right so we have what c d and f uh head back the forces that are known so we have maybe 20 kips here and we know that we're gonna have some reaction uh d y at a roller so there's going to be some vertical reaction add a roller and then what i like to do is i like to just always assume tension in trusses so that way if i get a negative value i know it's compression but what i like to do here is i'm just going to start in and i'm just going to draw my forces that i don't know so i'm going to draw ef i'll draw uh ce and i will draw bc so this is kind of a basic free body diagram that's kind of step one when you're getting into trusses so you know that's we're gonna get started with with this and then what we have to take a step back is okay so we're still trying to find this force this ce force so what i like to do whenever i get a question like this is to look for one of two things one is their parallel top and bottom chord because if so then i can just use sum of forces and find components if not what i like to do next is i'm just going to come back here and get a couple colors so what i what i like to do next is kind of look and see are there any points of intersection of the unknown forces that are going to be a convenient place to take a moment so in other words i'm going to draw a line right through ef that red line i just drew on the screen that's a line through the f that's kind of the line of action of ef that's the line that ef is passing through and that's where it's acting and similarly i like to draw one through bc and when you do that when you're doing method of sections this is actually a really cool thing because what you see all of a sudden is there's a point of intersection there and that point of intersection is an important point because that point is going to tell us where we can sum moments effectively so for this question what i'm going to do is i'm just going to sum my moments about point d okay so if i sum moments and this is just a sigma moment about point d equals zero and it's funny i was showing some of this stuff to my uh my sister-in-law who's a who's a algebra teacher she teaches algebra like all day every day she's like mark that looks like greek and it is greek uh there's a sigma but it's she 100 has the algebra she just it's like if and you guys you i know you have the algebra too the algebra is not the hard part the hard part is writing these equations and using these weird letters and putting them into into practice but all that we're doing here is the moment equation we're looking for a force times the distance and trying to figure out what forces cause moments so right away what we can see is there's a couple forces that don't pass through point d this this 20 kips and our force ce so those two forces have to balance each other rotationally in this in this free body diagram when we sum moments about point d any time i'm summing moments i can apply that force you know that force that unknown force like ce for example anywhere along its line of action in other words if i put a rope on that and pulled it somewhere i could put that anywhere along its line of action and it will cause the same moment so if i take that force for example right and i say well what's its line of action look like it looks like maybe this green line and you'll see that green line doesn't pass through point d and that's important because if i put that force ce anywhere along that line it's going to create the same moment about uh about point d anywhere along that line so so i can go anywhere along this line here and that force ce is going to create the same moment so what i like to do i mean if i knew what this perpendicular distance was that would be great if i knew that perpendicular distance i could solve for it directly i don't know that perpendicular since i could probably do some geometry and figure it out but but what i see here what i see is a little bit easier is i can just take that force and kind of bring it down to this point and now i can apply uh some components so i can apply like maybe c e y and c e x why is that important because what you see is c e x passes through point d so we're left with kind of one component here that we can solve directly so when i sum my moments i'm going to say well c e y times its moment arm so its moment arm is going to go from this point e to point d and if these are eight feet on center that's going to be a total of eight times two or sixteen feet right in which way is that which way is that causing rotation that's causing rotation kind of in this direction that matches our positive sign convention this is going to be a positive value okay and what else do we have we have we've plus 20 kips right so here's our 20 kips that's in a positive sign direction is what we're a positive rotation as well uh that's causing rotation in this direction and what's its moment are well it's the the force times the perpendicular distance the perpendicular distance from its line of action to point d is this horizontal distance of eight feet so i can put that in here this is times eight feet and all that has to equal zero those two forces have to balance each other and you'll notice i wrote ec up here in ce down here it's the same force and sorry for keeping any confusion on that but cec is the same force and now we can solve and we can solve specifically for c e y so we can see e y and what we'll get here is a negative value but if i put that in the calculator whatever i get i get 20 times 8 divided by 16 and that gives me a value of negative 10 kips she might be tempted to say okay we've got 10 kips i'm happy i'm just gonna move on but don't do that because that's just one component that's the vertical component and what we want are uh or what we want really is is this total force here so we want this total kind of resultant force here which is ce or ec however you want to do it so to look at that what we have to do next is kind of take a look at at this this whole triangle up here so if we take a look at this whole triangle right this whole triangle here is going to is going to help us to relate the component to the main force so the way i like to show that is by saying okay we know we have this triangle here right in this triangle between e f and c we have some dimensions it's the dimensions we have we know this is eight feet we know this is six feet and i'm a sucker for six eight and ten triangles so you'll have to forgive me for that but this is a six eight ten triangle okay and what that means is that's going to mimic or match our force triangle so if we have this force here and we have some components right and maybe i'll go back to saying this is ec just to uh just to be more consistent with what the problem says up here ec so ecy ecx and force all together ec right that's what we're trying to find here right so we're trying to find that ec and now this is just a similar triangle thing because i know that this angle and this angle are exactly the same so if we have similar triangles what we could do is well we could write well i know what do i know i know ec right and i know ecy so if i know those i can say well ec i don't know you see i'm sorry ec is the one i'm trying to find ec over ecy has to equal what it has to equal essentially 10 over over six right so when we have this here what we can see is all to get ec all we really have to do is is cross multiply and we get ec equal to 10 over 6 times e c y and when i did this out earlier i think i got like 16.67 right so 16.67 okay so the cool thing with this problem is we didn't even have to do that reaction i mean gut level first thing i do a lot of times when i see trusses like this is i'm going to well i'm going to go solve for d y but on the fe every minute that you can save is going to help you get to another problem where maybe you can't save it okay it's a little bit more difficult so if you can save not doing those reactions you know it saves you it saves you a few seconds there and that's going to help you but this should be our answer which fortunately is up here in our answers as well so i hope that one makes sense i mean it's just taking another look at trusses trying to introduce methodist sections a number of times here method of joints but hopefully this makes a little bit of sense and also hopefully it makes sense that we can apply right we can apply this moment anywhere i'm sorry we can apply this force anywhere along its line of action and it's going to equal the same moment and just again i don't want to try not to confuse people this can be ecx or cex it's the same same thing right it's it's it's not divided by east you know these are the same things i'm not trying to like mess with people with different notations here i probably should have kept this at ecx and ecy so hopefully that's not confusing but uh i don't know if you have questions definitely put them in the chat or we can uh you know keep going here but we'll just keep moving so uh this is good so let's go to question number two again we're in this analysis of statically determinate beams columns trusses and frames so this is you know we just did a truss now we're into this kind of beam frame type thing where we get the impact of two members connected by a pin and we have to look at what's one member going to do to the other member so one member is going to have an impact on the other member right so this this you know this top member here is going to have an impact on you know this member here so these are connected by a pin in the middle and what that means is in order to solve this question right this question asks for a moment reaction at point a right so down right down here point a the molar reaction at point a due to the applied loads is most nearly so what is it how do we get it well what we're going to do here is first we're going to separate this thing out and we realized well maybe we could take a look at this first but there's too many unknowns there's too many things that we don't know like if we if we were just to just draw a free body diagram of this thing you know we'd have a y and uh ax and cy and you can't forget this one right this is at a fixed support you have the moment at a if we're just at some moments at this point we'd have too many unknowns we'd have the moment at a and we'd have cy so we can't just do that directly and whenever you get problems like this that are uh you know separate or that have multiple members joined by pin that's where we have to separate them out so we're going to do here is we're just going to start with member bc so i'm going to address draw in member bc here and when i draw member bc right i've got this load that's two kips per foot and hopefully a uniformly loaded simply supported beam is something you've seen before right so this is one of those one of those ones that honestly when i was in consulting we'd interview people i would ask somebody some some people this question what's the maximum moment what's the maximum shear in this type of a beam a uniformly loaded simply supported beam it's one of those things if you know it's just w all over two it's really easy if you if even if you don't know that if you know the symmetry here right and this is 10 feet 2 times 10 is 20 kips right so we get a total kind of resultant force here of 20 kips if you know that hopefully you don't have to do a lot of work here to get that c y and b y are both going to equal 10 kips right that's the you know b y equals c y equals what w l over two we'll take a look at that a little bit later to the beam diagram if you don't remember what it is i mean again if you don't remember what it is that's okay you could also find it by just summing moments about point b equal to zero right if we did that we'd get you know minus 20 kips times the five feet so what's where's the five feet come from that's this five feet right so that's the moment arm uh and uh then we're gonna what we're gonna add cy times 10 feet so either way when we when we're all said and done we should get a value of cy equals well 20 divided by 2 is going to equal 10 kips so that's kind of the starting point once we have that uh what we have to do is we have to take a look at these pin reactions right this bx and by so i sort of left those over here for a second so we have b y and we have bx so when i separate this this piece at the pin what i'm doing is i'm kind of exposing these forces you'll notice i i didn't show the forces here to start because i didn't separate that piece apart but as soon as i separate it out that's when i have to show these forces that's when they kind of show up right the the force is balanced if the pin is connected once the pin is not connected they no longer balance and that's where we have to add them in so what i'm going to do next is i'm going to take a free body diagram of this other member here of a member a b so if i take a look at a b this is where i still have you know i still have my 21 kips right my 21 kips acting here yeah i'll get a one in there uh and this is at five feet from the support okay so i got my 21 kips but the question is what other what else do we have going on here we still know we have ax we still know we have a y we still know we have the moment at a and this is where equal and opposite side or sides of a pin will have equal and opposite reactions right so we have to deal with the ra those forces the bx and by forces that are in that pin right but when we show them you know up into the left or i'm sorry up and that's the right up into the right here right up into the right here on the other side of the pin we have to show them down so what's this b y and you know to the left bx so on opposite sides of pins the forces are equal and opposite so don't forget that we never solved this but we could come back up here and we could say well if we sum forces in the x direction equal to zero right put my sign convention in there just for for completeness but what we have here is we're just gonna get bx equals zero right so if if c is a roller bx has to be zero so that's just going to be zero as well so what we know here is i'm just going to take this and i'll leave it in there for now i'll just call it i'll call it zero right so this equals zero and b y equals what do we get we got b y to equal to 10 kips so now this is just a sum of moments equation and that's what we're trying to do we're trying to find the moment reaction at point a so now what i can do here is i can sum moments at point a and if i do that i'm just gonna sum my moments at point a make that equal to zero and what do i get i get you have to remember your mother uh you have to remember ma right your mother your mother your well the moment of day is a moment today you got to remember that one it's there don't forget about it and honestly when you're done with this go give your mom a ring and just say hey mom thanks i i love you i'm so glad you you believe the best of me sorry that was just an aside that has nothing to do with the fe but um it's just a piece of advice you know so as we go through there what we're gonna do is we're gonna some moments at point a well forces don't pass through point a again we have some couple forces here we have the 21 kips and we have b y in a moment is what a moment [Music] is equal to a force times a perpendicular distance so the cool thing here is the 21 kips i probably could have put this in the problem statement you'll probably see that this is perpendicular but i'll tell you now that force is perpendicular to member a b and if that's perpendicular the cool thing is we already have this moment arm we already have the perpendicular distance so the moment of day is going to be 21 kips times i'm sorry the moment of a is going to be we're going to subtract off 21 kips times this 5 feet and then the only other force here is this by and we have to figure out its moment arm so what i'm going to do is if i kind of extend the line here its line of action is going to have a moment arm a perpendicular moment or from point a to the line of action of well ah shoot we don't even have this right we know six and ten and there's my favorite one again right that favorite triangle of mine where we have a six 8 10 triangle so 3 4 5 6 8 10 i'd have to think to to go even higher but uh it's it's some of these pythagorean theorem type triangles if you have you know the the hypotenuse in one side you should be able to find the other okay so a squared plus b squared equals c squared type of thing okay so that's going to give us our moment times our force times the perpendicular distance is going to give us our moments we've got the minus 1 times 5 and then we also know that this uh distance here is that eight feet so what that what does that mean uh we're gonna subtract off as well b y times eight feet that's gonna equal zero so the good news is we we basically have one equation one unknown and if we're summing moments about point a that those are the only two things that cause a moment about point a so we get the moment at a uh equal to what 21 times five uh you know plus b y which is 10 times eight so if i do that on my calculator hopefully we get an answer but i i think we do so we get 21 times five i think that's 105 yeah so 105 kip feet plus 80 not kip feet so that's going to equal a 185 uh kip beat okay and that's our answer right so that's the answer that we're looking for here but with with frames what you're gonna have to do is you're gonna need to separate them probably at that internal pin so that internal pin is something you have to deal with uh and then what you're gonna have to deal with is you're gonna have to deal with the reactions on each side of it so on each side of a pin they're gonna be equal and opposite so when you put that pin back together those forces cancel out when you pull them apart you see them okay so take a look at that you know that's that's one of the things that you're going to have to to deal with all right okay well let's keep going so we've got two questions down i mean honestly today's today's session probably won't be as long as some of the last couple so hopefully we'll uh get to bed a little earlier who knows but let's look at this deflection of statically determinate beams columns trusses and frames so now we're at a deflection in the deflection there's a couple different methods i mean some of you probably learned how to integrate and integrate and integrate and integrate until you finally get a deflection formula and you can do that and and when you get questions like this that are piecewise they get annoying honestly they just get long and annoying and there's an easier way okay so the easier way is to come in to your reference handbook here and in the reference handbook if we go down to civil engineering there should be a structural section right so we go past geotechnical and eventually we get to the structures section so this is kind of helpful uh except that i think i'm looking in the wrong section and it's less helpful there so we have a couple things we have a static indeterminate you know we've fixed end moments your structural design and we think uh oh we missed it where are those beam tables those beam tables are actually in mechanics and materials i think so i think i got that wrong but let's go back here and look and honestly there are going to be some pieces in structural analysis that go back in in mechanics and material so like i talked about earlier like the the column formulas the columns columns show up in oh that's what we're looking for we'll come back to that but well we'll come back to it let's just do it now so so what we see here is we have some beam diagrams right and these beam diagrams are going to be what we're going to use and i just picked one of them i mean it could it could show up like any of them this is one of those super common ones that simply supported uniformly loaded beam i've wl to the fourth over 3d4 ui the maximum moments wl squared over 8. that's not the one we're using but i pointed it out just because it's a super uh it's a super common one you know and you might be tempted to look at this and say oh that's the same loading we have and it's half the span but don't use it because it has a pin on one side and a roller on the other right so what we're doing is we're going to come down here and we're going to look at where we have a cantilever beam and we have this this this maximum deflection at its end so that's what the question asked for right so if we come back to the question we see we have a steel flange section better out of strong axis it's originally connected at support a so rigid connection means it's resisting rotation this is a cantilever and the magnitude the maximum deflection so if this thing is going to deflect it's going to deflect kind of like this and we're looking for this kind of i like to use delta this is going to be like our maximum deflection the the reference handbook use that v or new you know but but i just like to use this so what does that mean so let's come back here and see what else we got so we got this formula here we got a maximum deflection and we're gonna get 7wl to the fourth over 384ei so let's just write that out and so so the formula that we're solving here is 7 so our delta max or our new max is going to equal 7wl to the fourth over 384 ei and i dropped the negative sign because i don't really care so much that this is going down i mean i know the load is pushing it down but you could keep the negative if you want um so let's go let's go with it so let's let's go here okay so i just saw a comment and so i'm gonna hit that comment real quick how come the first two questions aren't just considered statics they kind of are they kind of are statics okay uh and just not to burst your bubble but they kind of are statics but when we're looking at it okay i'm just gonna come back here for one second when we're looking at analysis a statically determinant structures typically we're just looking at statics it's just maybe a little bit more complicated than it was in your 200 level statics and now this is 300 level and it's a little bit more complicated it's sort of like pre-algebra and algebra you're doing the same things or pre-calculus and calculus you're kind of doing a lot of the same things but sometimes it's a little bit harder in calculus versus pre-calculus static versus structural analysis okay that's kind of the the way i'll throw it there so sorry for seeing that coming late but let's keep going here so let's define our variables right so our variables here we have our w and that's going to be our uniform load that's just going to be two kips per foot okay good uh l well as l 8 feeders at 16 feet now let's go back and take a look and see what that is so if we come back and we look at our reference handbook here uh what do we get we get l if we look at l we see l over 2 is going to be half the span and l is going to be the whole span so let's come back here and look at l is going to equal the whole span 16 feet uh e [Music] do you know e is off the top of your head uh for steel beam hopefully you you've memorized e if not there was that table i think we just passed over it right we looked at this maybe in the mechanics material section but what do we get right so for e uh if we come down for e steel we get 29 million psi so 29 mpsi so what do we have we have 29 m psi which is equivalent to 29 000 uh ksi or i mean we could say with this is also equal to 29 million psi right so you have to know your unit conversions there we got some unit conversions and then we're going to look at i and some of you are going to be stumped and saying like hey isn't there a steel manual for that and the answer is yes there is and the other answer is some of these members are actually shown in the the civil engineering section so if we come back and look at the civil engineering section here right and i come in and i look and i scroll down far enough i should be able to find we'll come back to geotech in a couple weeks that'll be fun um i should be able to find some sections so you could just do a quick control f that's probably going to get me there sooner oh man i just probably should have done it so control f would have gotten me there sooner right an 18 by 71 though is i get some values here so if i look at that i'm going to zoom in just a little bit here so what i'm looking for is an 18 by 71 i got that and i'm looking for eyes so i see i here and then i hear which one day do i use if it's the strong axis you use the big ones so the strong axis is going to be the x value so we're going to take the w 18 by 71 i value here of what's that 11 70. okay so 11 70 and that's in inches to the fourth did i say units are important um maybe not this one this this session so far but units are super important so let's look at our units here as we plug this in so seven times w which is two kips per foot times our length which is 16 feet that's to the fourth okay and then i'm going to divide that by 384 e which is in since i have kips up on the top i'm going to use kips on the bottom i'm going to come back in here and i'm going to say this is going to be times 29 000. kips per square inch sometimes i like it like rather than write ksi i like to write kips per square inch like that because it's easier to to to cancel out units right so we have the kips per square inch and what else we have we have our i our i was 11 70 inches to the fourth okay so we have all the numbers but you'll notice that we have we have units that are inconsistent we have feet on the top we have inches on the bottom the kips are nice because the kips cancel out okay so the kips cancel out those are good but if we just look at units we've got we've got feet to the fourth divided by feet right that's our that's our feet to the fourth divided by feet and then on the bottom we have you know inches squared uh under well here let me actually let me write it this way we've got inches to the 4th divided by inches squared so if we keep going we're going to get feet cubed divided by inches squared and we see we have to do a conversion so the conversion that i'm going to do is i'm going to multiply this by what i know for every one foot i have my units are in inches so i'm going to try to get this to inches for every one foot i have i have 12 inches and i'm going to cube that when i cube that what happens is this the feet go away um all you know all the inches except for one this kind of gets me to units of inches right so the red here is just the the units the the red is just the units but basically you what the the long and the short of this is this is these are kind of like that's not equal these are just units right so this is just units over here and if i could spell it that would even be more helpful so units okay so what we have here is these are these are all our units but we need to multiply this by 12 cubed in order to get our answer so once you know the units this gets a little bit easier this is a little bit of a maze to go look up the different pieces whether it's e or it's i or it's the formula in the first place so you have to be really familiar with that reference handbook to go do this but if we type this in now hopefully we get the right answer so 7 times 2 times 16 to the fourth right times 12 to the third okay and then divide all that by 384 and divide that by 29 000 i can plug it in right here and divide that by 11 70 which is our moment of inertia and i get like 0.12 so 0.12 uh one two it's like one two one or one two two um inches so that's close enough it's most nearly the 0.12 okay but this this brings up that approach the fe where you have to understand you have to totally understand where these formulas are when they will look them up you have to understand what the units are you have to understand kind of where to find uh your your e your modulus elasticity if you don't just remember it and you have to know how to find this i value because they're it's not just going to be given i mean maybe it'll be given to you if they do give it to you that's even better then you don't have to go find it but there it is all right question three is done okay let's keep going because we got another deflection one and for those of you that took structural analysis with me you guys are going to grow and you're like principle of virtual work oh man that's terrible okay it can be terrible it can be it can be complicated but i've you know said i've been surprised sometimes talking to people where where you know they'll take fe concepts that i think are too complicated to throw on the example they'll take them and throw them on the exam and try to make them less complicated okay so uh so we'll go from there so so let's take a look at this one right so ah man i keep seeing comments a little bit late here i'll try to keep a better eye on it but thanks mikey for jumping in there the x-axis right if we're if we're bent about its strong axis that's i x the strong axis is i x there you go strong axis is ix always uh for for what we're doing in this this uh for this test let's come back here though principal virtual work for trusses deflection and trust as they say one of the one of the requirements here is that deflection is statically determinant trusses okay we kind of did columns last time with that pl over ae formula if you remember the pl over ae formula with a couple members that in mechanics materials section and we said well what force will cause this thing to deflect a certain amount so that's kind of like a column of pl over ae and the trusses is honestly very similar you got a lot of little columns connected by joints so let's go to the principal order work here and see what that looks like so if we come back to our our reference handbook here what we'll see here is actually i'm going to keep coming back past steel components back into our structural analysis section so if i come back all the way to structural analysis section uh there's determinancy we'll come back to determinacy in a little bit here but i want to just keep coming here and this is frame deflection using the unit load method again this is more kind of uh this is more calculus integrations this is kind of a a follow up to the question we're doing so this is a little bit more complicated than the question we're doing here but the question we're doing here is going to take a look at the trust deflection by the unit load method so this is a similar type of thing but instead of using integrate integrals we're using a summation it makes it a little bit easier makes it a little bit more doable but what we see here is a big formula and you might see this formula and just say flag i'm moving on and uh not even not even trying to solve it but honestly if you see a principle virtual work uh problem for a trust or deflection in a trust or something like this it's probably going to be actually a fairly simple trust or a fairly easy problem in the sense that there's going to be a lot of zero force members to make the analysis doable in a short amount of time so what we see here is this is this uh is this formula of we have a delta of a joint so displacement of a joint is equal to the summation of f i which is the force in the member caused by a unit load i'm like a one kilonewton load here times delta l i so delta i remember that p l over a formula from the last time this is f l over a e it's the same thing but f is the force instead of p right so we have this f times f val over a e okay so the um so so yeah i mean this kind of gets in there principal virtue works a little it's it's it works kind of the same way um but yeah we're not going to get into all that there is i do does this have a principal virtual work section i'm trying to think uh if i remember in here correctly we i don't remember i'd have to search um but why not why don't i burr jewel uh work sorry i can't spell today actually my keyboard is off to the side so there is no principal version this is essentially a principle of virtual work okay so it's just that that's where we're going with it okay so um so let's just jump in here and we'll we'll take a look at what this looks like okay so so so let me come back over let me come back over here um what we have is a formula that essentially says the delta equals the sum of little f you know big f l over a e okay so we need to sum that so we need to find a couple things and this is where the little f is the force based on a unit load big f is the force based on the actual loads okay uh the l is the length of the member a is the area and e is uh the modulus of elasticity all right so with this we have to find a couple things but what we can see here is is right off the right of the back we see well if we put what are we trying to find we're trying to find we have well we have a trust we're given five centimeter by five centimeters square members with a modulus elasticity of two hundred thousand so this question the modulus less says cities just given okay ignore the self-weight trust ignore self-weight assume compression members are braced against buckling so we're not getting weird things going on and you don't have to evaluate it for buckling um the magnitude the horizontal displacement at point d due to the applied loads is most nearly so any time we're doing this what we're going to do is we're going to put like a unit load at that point so we're going to put this like one kilonewton load at that point and and first we're going to analyze it like that and if we do that i'm just gonna throw through a couple things out there real quick so do you remember zero force members zero force members could totally show up on the exam as well but but zero force members here if if you remember like let's just let's just take a walk down zero force member lane for a minute so if we look at like joint c for example okay joint c um we have a couple forces coming out of joint c well this would be like cd and like ac if you just look at joint c um basically what we have going on here is we've got joint c in and if you take a look at this um we're looking to see that the sum of the forces in the y direction equals zero the sum of the forces in the x direction equals zero we can see from the y that ac equals zero we can see from the x that c d also equals zero okay so we can see both of those right so so hopefully that makes sense and isn't too crazy but what that tells us here is that this is zero force this is a zero force similarly if we look at you know joint b at the bottom here if we look at joint b all right if we look at joint b and this is what joint uh b um we're going to have some reaction at b right this is going to be like b y i will probably have like a b d and then here we're gonna have like a b but if you remember i mean hopefully you don't have to go through these free body diagrams on the fe i'm doing this here just to kind of refresh your memory about zero force members right do you remember zero force members zero force uh members right so this is where hopefully you can remember some of these things but what you can see here is if we sum forces in the x direction equals zero um a b equals zero okay so so basically same thing going on here and what does that mean so that means we have a force here that's zero so there's only two forces in this truss that are gonna have to you know have to hold or i'm sorry there are only two members in this trust that are gonna have to hold four so we have uh a d and db okay so hopefully just from remembering a few things about zero force members you can go ahead and you can take a look at these members and just say oh you know not having to do this whole kind of like analysis here um you can go ahead and solve and say okay i know those are zero force members i really only need to solve for these two forces so if we do that what we're going to do here is we're going to take that joint what joint do we have left we've joined let's look at joint d and i'm going to put my one kilonewton force here and i'm going gonna have my force what's this this is maybe called a d and force uh a b i'm gonna put a b up because i kind of know it's in compression but let's take a look at the geometry here and see if you guys can do this maybe in your head a little bit even but if we take a look at this thing going on what do we have we have the sum of the forces in the x direction equals zero right so what do we get we get this component a dx has to equal essentially adx has to equal one kilonewton do you see it these two forces have to be equal horizontally and then this truss got even easier because we have a 4 4 triangle so i tried to make this kind of easy so the math wasn't too crazy you didn't have to go do geometry to relate components um here hopefully you can see um since it's a you know a 45 degree you know triangle that means that means what it means a d y also equals one kilonewton okay i'll get my head out of the way um right i mean can you just can you can you do that without having having to write it down i hope so and hopefully that means now once we have that we can say some of the forces in the y direction equals zero and we can say well a b you know a b equals a d y equals one kilonewton right so this trust the the analysis of this trust everything just equals one i mean it's it's kind of a simplified version here but hopefully it makes a little bit of sense right i mean basically the horizontal forces have to balance so the one killing balance is the adx the the vertical and horizontal components of a d have to be the same i should probably label this one here a d why they have to be the same so since they have to be the same that means you know they both have to be one then a b and a d y they're the only two vertical forces so they have to be one so again truss analysis but we're taking it a little bit further here okay the last thing that i want to do is with that 40 feet five degree triangle we also can solve for a d and that's just going to equal this the square root of two uh kilonewtons so one times the square root of two okay once we have that okay once we have that we can take a look up here and what we see is what's the difference between the free body diagram we just drew and the so this is for like little forces f i right so that's for little forces f i what's the difference between that one and you know if we have a d 70 kilonewtons and a b you see the only difference is multiplied by 70. right so once you solve it once for one of these as long as all there's only one load it makes it really easy if there's only one load it's it's super easy and i say super easy you don't have to go reanalyze it so this you know this is essentially if we have f i capital f i right so this is the this is the big oops this is a big load here right f i is just going to equal 70 times little f i is that okay and because it's all the same geometry and it's all the same except for the the magnitude of the load okay so if we're if we're doing the unit load the principal virtual work here for trust is what are we going to do we're going to say that we i'm just going to make a table i'm going to say fi f i uh and then what else do we have we have what's our formula it's f l over a e so i'm just gonna do ffl and then f f l so i'm just gonna i'm gonna leave a e out for a second and what i'm gonna do here is i'm gonna do one for each member and i'm ah that didn't look good so i'm just gonna rewrite it here for a second f capital f l so what members do we have we have two members basically we have member members i'll just write it here a d and a b right so we have two members a d we said little f i right so little f i a d was going to be the square root of 2 and big f is going to be 70 square root of 2. and then the length right the length if we have a 4 4 triangle this length is going to be 4 square root of 2. okay so let's come down here so 4 square root of 2. and if we multiply all that out what do we get we get 70 times 4 times two times the square root of two did i do that right i got like 791 or 792 roughly okay i multiply all that out a b a b what's a b well a b we said was one we said it was 70. we said the the the height here that height is going to be just the four meters right so what's the height it's gonna be uh four so four times seventy what's that 280 i i believe so if i can still do math that so if i add those two together 792 uh plus 480 or 280 i'm sorry i get about 10 72 for ffl and somebody's out there is wondering well what about a and e well a and e are constant so i'm not going to put those into my my sum here i'm going to i'm going to do those afterwards so let's do this out and we'll we'll plug in the equation so so here's kind of where that principal virtual work comes in we're doing 1 times delta is going to equal 1 kilo newton times delta equals what it equals the sum of f f l over a e and since ae is constant we can pull this out so now i'm just going to write this in so what's that sum this sum is right here this is going to be 10 72 and if you look at the units i'm going to have kilonewtons squared times meters and now i'm going to divide and what am i dividing by well i'm given centimeters and megapascals so what units you want to use the units that i'm going to use here are i'm going to say well i'm going to say i'm just going to translate this to millimeters and i'm going to say 50 millimeters times 50 millimeters times 200 000 megapascals megapascals is a newton per millimeter squared okay so let's see if i can do my units right and uh so what do i need to do here i need to i need to figure a couple units out so let me just make sure i'm doing my units right because i think i screwed something up did i screw something up maybe i might have screwed up units here so let's just take a look let's see what we got um the kilonewtons go away right so if i'm just going to cross things out the kilonewtons go away i get one kilogram to go away uh what else do i need to do well now i have the millimeter squared go away and i've got some inconsistencies with the main unit so i've got a killing on the top of newton on the bottom i've got a meter on the top and if we're looking here we want our deflection or displacement in millimeters so i want to convert everything to millimeters so i'm going to come back down here and what i want to do here is i want to get rid of that kilonewton so i'm going to multiply times for every um 1000 newtons i'm going to have one kilonewton so my kilonewtons are going to cancel out my newtons are going to cancel out and then what else do i have i have 1 000 millimeters per 1 meter okay so is this a long problem yeah it is it's a long problem but it hits on some fundamental principles right the fundamental principles are like units do you know units do you know how to apply some of these more complicated methods in structural analysis and can you put it into your calculator and take the basic trust analysis and apply uh something that's a little bit more complicated here so if we type 1072 times a thousand squared you know divide by 50 divided by 50 divided by 200 thousand i get 2.144 millimeters so i get my delta equal to 2.14 millimeters which should match one of these up here so in theory this matches our two millimeters it's most closely related to that okay so is this probably going to be on there i don't know maybe maybe not could i see a problem like this where instead of having the load be you know horizontal if the load is maybe vertical i could see that why why is that that that question's even easier because when you look at this question this goes to zero this is zero this is your they're all zero force members except for one member here and then this just simplifies to what it simplifies to like a p l over a e problem so like column right i mean it's that's where it's like it could get it could get simpler right and this isn't the problem we have you know this isn't the problem that that um we're given you know in this problem statement but is it something that fe could give you sure why not it's that same idea of like i'm gonna give you something that looks really complicated but once everything goes to zero it makes it a little bit less complicated a little bit more doable in that five minutes but if you understand the fundamentals of the trust analysis of zero force members this that's gonna help you a ton oh man that was that was um quite the problem you know but let's just let's keep going if you have questions let me know but hopefully this is helpful as well just remember in zero force members remember that like you know principle virtual work kind of thing the the the unit load uh type of thing so let's keep moving so we are on to question five and this goes into column analysis buckling boundary conditions uh probably when we get into steel analysis we'll get more into columns as well but this one kind of just shows it shows up and says with a bar it's used as a conjures as axial forces and compression now the column is solid steel so we get we get that um you know e thinking in the back of our head what's e um for centimeters it's going to be that 200 okay they give it to us this is cross section of two centimeters by six centimeters and modulus elasticity 2200 it's simply supported at its end for both the strong and weak axis so we have simple supports at both ends okay and then in the weak axis it's braced at the midpoint as well okay only in the weak axis and not in the strong axis okay so what's the theoretical euler buckling load that'll cause the bar to buckle okay so this is where you kind of need to know what those uh those those those formulas all right so if we come back and this is where this is a great one for the for uh control f so if i just start to type in buckle there's lateral torsional buckling uh and honestly i'm gonna go to previous and look here so so if we have columns uh if you kind of know how this this works we have columns we get a critical axial load for long column subject to buckling euler's formulas p critical pi squared ei over kl squared the other thing i'll throw out here is you get pi squared e i i'm sorry pi squared e over k l over r squared sometimes that k l over r it's also known as the slenderness ratio you have to know what r is r is the radius of gyration it's just the square root of i over a so i might be asked as well like it's radius of gyration you might be given a critical buckling stress instead of a critical blocking load you kind of have to use one equation or the other but let's go here and we will take a look at that so we have pcr's pi squared e i over kl squared okay so we'll come back here and we get oh i'm just going to write that down pcr equals pi squared ei over k l squared and the nice thing for you is when you're taking the fe you're gonna have the handbook on one side you're gonna have your work kind of in front of you on your little you know tablet thing that you get to write on and then you'll have your problems on the other side that you get to work through as well so this is our pcr so so we we already we're given e right so we're given uh we're given e so e we know is 200 gpa uh we don't know k we have to go look up k we kind of have these l values here that we can use and specifically l is going to change depending on which axis you're at so if we're looking at like the weak axis let me uh let's let's let's just draw it down here why not and so if we're looking at the weak axis right we know that the unbraced length is lcy the unbraced length you can call it i'll see why you call it l y it's probably l y is a better uh term for it but l y the unbraced length is going to be 1.5 meters right the length between bracing points is just that 1.5 meters similarly uh lx right so we have lx i'm sorry actually let me come over to let me let me come back here for a second and i'm just going to do the strong axis on this side okay so similarly over here the strong axis we have lx and this is where we get that total length we get the total length which is three meters because it's it's it's braced at three meters okay we need to think about k y and k x right and this is where hopefully you kind of remember what that is but this is kind of that uh effective length factor so if we're looking at how these things are going to buckle we're going to have this buckling and kind of a bow shape right so we're gonna have a bow shape here but but for the weak axis we're gonna bow but it's gonna it's gonna come back on itself here so it's gonna go through that bracing point so for the the strong axis there is no extra bracing point it's just going to be braced right at both of the pins right so but for the weak axis it's going to be braced at these these three points here so we have to look up the effective length factors and if we come back to our reference handbook let's see if we have them let's see if we if we can find them here so um you know if we look at this we have k which uh my computer is not liking me today so let's let's uh see if i can fix that uh or maybe it's just okay i don't know the k what we have here is theoretical effective length factors pin pin is gonna be k equal to one fixed pin point five or fixed fix six point five fixed pin you know point 0.72 but effectively what we have here is we have pinned pinned for both of them right so if if if we think of a pin what's a pin pin allows free rotation it allows free rotation so if it this point and at this point we have free rotation so we're going to say k equal to uh 1.0 uh similarly over here we have free rotation at both ends and kx is going to be 1.0 we got the k we got the l we know that e already we said e was well e was 200 uh gpa which is equal to 200 000 mega pascals okay however whatever units you like and and then what else do we need we need i so let's come to strong axis and weak axis i equals you know i y this goes back to that question that came up kind of earlier what's strong axis what's weak axis well the strong is the one where essentially get the bigger h value right so it's like it depends on how you're going to buckle this thing but the bigger h value so if we come back to our our mechanics materials section in here or well actually this is civil engineering now but if we come back to our mechanics materials section we have some values here for uh for a moment of inertia and we kind of went over this in the mechanics material section but it's worth doing again because it shows up uh when we get to actually it's in the static section oh man i missed that one this is in static section moment of inertia is in the static centroids uh centroids masses areas lengths and volumes right this is where we have a standard value here and we have i x is base times height cubed so the big height here versus i y is the base cubed times the height okay so in other words when we come back here we're going to see the general rectangle formula here is base height i equals base height cubed over 12. so when we go to the weak axis we want the smaller moment of inertia so what's this going to be it's going to be 6 times 2 cubed over 12. and honestly i'm going to i'm just going to make the commitment now i'm just going to make the commitment and i'm going to say rather than do that i'm going to say if this is what do i have if two centimeters by six centimeters i'm going to say um i'm sorry i'm going to say i'm going to just convert everything to millimeters i i just i'm going to say 60 millimeters times 20 millimeters cubed and that's going to give me this value here right so 60 times 20 to the 3 over 12 uh what do i get i get about 40 000 and so 40 000 uh millimeters to the fourth okay and then we can do the same thing for the strong axis but for the strong axis the h is going to be the one that makes it bigger so what i'm going to do here i'm going to say this equals to 20 millimeters times 60 millimeters cubed all over at 12 and that's going to be a bigger number because the 60 gets cubed not the the 20 right so uh so what i'm going to find here i'm going to go to 60 and i'm going to get 360 000 millimeters to the fourth okay so that gives us that gives us some things and then all we have to do really is is we have all our values we just need to get the right units and and use or apply our equation right so we're going to get pcry and pcrx i'm just going to make a little bit of room here and i'm going to say well p c r y the critical buckling load for y is going to be pi squared e which is going to be two hundred thousand i'm just going to do mega pascals or newtons per uh meters squared times what was it there's pi squared e i over kl squared so times um i which is 40 uh thousand millimeters to the fourth okay and then we're gonna divide that by kl squared so we're gonna have 1.5 or 1500 millimeters to the fourth i'm doing a conversion on the fly there but 1500 millimeters to the fourth okay so what do we get well let's plug it in because i don't remember so pi squared uh times 20 no not 20 200 000 high square times 200 000 times 40 000. so big numbers but it's either really big numbers or really small numbers of the meters i like the big numbers but some people like them smaller i don't know and so what do i get like um something that doesn't make sense so what did i do wrong uh oh um i don't know i probably did something wrong oh i'm sorry it's not 1500 to the fourth it's over kl squared there it is it's a kl squared i was getting too many force going on in here that's a square um it also helps on the fe when you write down a formula to actually use that formula just saying it's going to help so the nice thing with your calculator is if you get a weird number you can just go back to second enter and and sort of change it around here and you know and then i get the 35 000. that makes more sense so pcry equals about 35 you know 100 newtons 35 100 okay so we could just stop and say great we got 35 but what if what if the strong axis because it's got that longer three meter length what if the strong axis actually buckles in ten kilonewtons then we'd feel really sad because we'd get the problem wrong so let's go and get that that's just i mean it doesn't take that much longer to do this all right let's do pcrx and just confirm that that's not the 10 you know 10 kilonewtons and uh be on our way so pi squared times 200 000 newtons per millimeter squared times 360 1000 millimeters to the fourth divided by that kl squared and we'll get it right this time so 3000 millimeters squared so pcrx is this going to be smaller than pcry if it is it'll buckle that way which which one is it so um the nice thing is with your calculator you can do is you can just go back to the previous one that you just did and modify them or you just plug it in all again i mean it's it's up to you how you like to use your your calculator but this is where it's like it's helpful to get a calculator that you like and do what i call um calculator practice just plug these things in and see if you can get the same the same values because what i get here is is roughly 79 000 newtons okay and conceivably you could have a problem where you have you know more weak axis bracing and the strong axis actually controls or you could have something where maybe the x and the y are closer and the strong axis controls as well so it's not going to be a given that the the weak access controls most of the time that's what it's going to be i but but for here you know the the value of 35 000 newtons or 35 kilonewtons is what's going to happen okay so i mean if you had a smaller h maybe the strong axis would control it's just it depends you know on the on the values that you're given but this is where understanding kind of this value understanding some of these values is going to be helpful so you got l you got k you got the i you got pcr and then again just just to throw this one out there because this can show up too if you get the buckling stress so stress um stress is dependent on kl over r which is your slenderness ratio and you just have to remember what's our r equals the square root of i over a that's the radius of gyration and it i mean for for like the weak axis for example what would this be just be oh we would have we'd have 40 000 millimeters to the fourth divided by its area of what 60 millimeters times 20 millimeters and take the square root of all that you might be saying matson why are you doing that because um didn't we just already solve this problem yeah this is just extra this is just like extra this is like a bonus and the reason you know the extra and i'm just gonna say you know needed uh for buckling stress right and the other reason it might be needed is because i was talking to somebody they're like yeah they asked for radius of gyration man they so so what's radius of your iteration it's the square to i array okay so the question is why did i pick the uh um why did i pick the the the weak axis over the strong axis in this case why to pick the 35 kilonewtons versus the 79 kilonewtons before i get there i'm just going to do this math out real quick so if i do the square root of 40 000 divided by 60 divided by 20 i mean a value of like a radius iteration of like 5.77 inches okay so that's just a little tidbit you you need that value if you're doing the buckling stress so i just wanted to throw that one out there so you have it nothing you need it for this one so why did i pick pcry or pcrx so let's just come back up here for a second think about this as you're loading this thing this column gets one axial load it gets one axial load p so the question is when is it going to buckle it's gonna buckle when i mean if we're taking a look at papers right it's going to buckle like do you see this this is going to buckle eventually and it's never going to buckle on a strong axis with a piece of paper it's just going to buckle on the weak axis right but if we brace in the middle it's not a piece of paper so thin it's still gonna buckle on that weak axis it's conceivable if you brace this one in the weak axis it's conceivable that this strong axis could buckle first but if we think about it like this we have one common load p that's being pushed down here if the buckling load for you know this side is 35 kilonewtons the buckling load here is 79 kilonewtons we're never going to make 79 kilonewtons because at 35 it's going to buckle does that make sense oh it's already going to be buckled before you can even make it to the 79 so the 35 is going to control i hope that answers the question uh but you know if not definitely put some more things out there i would love to try to answer if i can okay the smallest load perfectly i love that answer smallest load that will cause buckling okay awesome oh man let's go back to structural determinancy and we'll just keep moving on here structural determinancy stability analysis of beams trusses frames i mean we're on question six we thought we're to get done early um hopefully we can still get done early here but let's look at this we were just talking too much doing too many extra things like radius of gyration structural determinacy so if i just search determine c right can i did i do the right word um indeterminate indeterminacy uh what else do i get determine classifications of structures so this is interesting because i give you you know this is one of those things i sort of throw into my structural analysis class just to kind of get the idea of what determinancy is it's it's not something you necessarily always do when you're working in the real world but it's one of those things that you have to understand whether structure's stable it's i'm saying it's unstable it's stable and determinate in other words you can determine everything just by using the equations of equilibrium or it's indeterminate and you can see throw up your hand and say i can't determine it get it um so so i can't determine it in the sense that the equations of equilibrium aren't enough to determine everything so we can do this with trusses we can do it with frames and what normally that i like to do here is is i sort of take this a step back i mean three times the number joints plus c all right so 3m plus r um normally i think of it a little bit differently than that so oh man let's just take a look back over here normally when i'm looking at structural determinancy what i know is i have three equations of equilibrium some force is x some forces y and sum of moments for every beam if i have three unknowns i can solve that beam just using those three equations if i don't have those beams it's an indeterminate structure and it's gonna be harder you're gonna have to use some other method okay but what i like to think of is is specifically here i'll just write out the notes again here for a second but if if i have basically what i'm gonna do is i'm gonna count the number of beams so i'm going to say you know i'm going to look at the reactions so if i if i look at the number of reactions and if that is equal to three times the number of beams it's going to be statically determinate as long as it's stable so this question says which of the structures below are both stat stable and statically determined this might be one of those alternative types of questions where you just have to click and you have to click ah which ones do i click there are six of them i can't just pick a and this is this is one of those annoying ones where it's like you don't know and you can't just pick b because b is the answer that i like to use the most i don't know you know it's like you can't pick c sees the the the uh the the common answer i don't know so what you have to do is you have to kind of understand so when i see these structures what i do is i like to just sort of draw in reaction forces instead of support symbols so i'm going to draw in one reaction force here for that roller i'm going to draw in two forces at the pin because remember those forces on either side of the pin are equal and opposite i'm going to draw in another reaction force at a roller and two more uh at a pin i like to draw the ones at the pin as vertical and horizontal normally but then what i like to do is just go back and count them up one two three four five six so i've got six um six reactions so this is kind of like if we think of it like this six reaction 6r and that is is that equal to or greater than or less than three times the number of beams so i've got one beam here and two second beam here so i've got three times two beams equals six so this structure is technically statically determined you can determine every support reaction pin reaction just by using the equations of equilibrium so that checks this structure is statically determinant and you would circle this or put this down as statically determined okay good so now that we've done that let's let's apply our algorithm to the same thing we've got a roller we got a roller we got a roller great so we can count these up we got one two three so we got you know we got one beam okay so this is great we got we got what we got three reactions is that equal to three times our one beam yes so three times three equals three you know whatever three equals three what's the problem the problem is any force what do rollers do the problem is rollers roll things right the problem is any force on this thing the horizontal direction is going to make this beam just roll away and even though we got three equals three this is just means this one is going to be what this is bad because this is unstable right there's nothing resisting the horizontal forces this is unstable and it's a big fat no okay it's not stable and steadily determined okay let's go to the next one and again i just like to sort of apply this so i got one two three one two three so every fixed reaction is gonna have three how do you prove it you don't you just have to look at it i mean it's like this is yeah this isn't like a prove or unproof thing it's just a thing thing um so we got one two you know one two and uh what do we got one two three actually there is a way to prove it uh uh maybe i'll show that to you in a second hold on four five six seven eight nine ten anytime we don't get a multiple of three we should just stop and think for a second but here how many beams we have we have one two three beams so what do we get we get how many reactions we have 10 reactions we got we got three times well actually um ten reactions and three times three is nine you know nine equations and ten is bigger than nine so that means this is statically indeterminate to the first degree okay so that i just write s i to the first degree but also that means this is not statically determinate okay how do we prove this one here one of the things is if you sum forces i mean honestly if if you sum forces in the x direction and let's say we put this force p here we sum forces in the x direction equals zero p does not equal zero we know that p is some value and p doesn't equal zero p is the only force in the x direction it's not being resisted by anything so that means it's going to move there's going to be some mass times acceleration it's not static anymore that makes sense so that's one way i'd say you could prove that one okay let's keep going here so we got a roller a pin so we're rocking and rolling here one you know we got rocking and rolling and we're good so um let's just count these up so one two three four five six seven okay seven did i count all those right no i missed one i missed one in here i got i got an eighth one in here that i missed i got a two at the pen okay so i got eight reactions eight uh actually let me just switch colors here so i've got eight reactions maybe um eight reactions which is less than uh how many beams we have we have one two three beams less than nine that means this is just unstable you don't have to do anything else we don't have enough reactions i mean honestly you can just look at this one and hopefully see that this thing is unstable right because this pin this uh this thing is just going to want to collapse it's a collapsible mechanism okay so if you have got a collapsible mechanism you don't even have to keep going because it's just going to collapse and that's unstable so let's keep going though and figure out what we're doing next we got a roller we've got a pin a roller has one a pin has two a fixed connection has three reactions so uh we can do the same thing kind of going here we got one two three four five six this is looking good we got a multiple of three how many beams do we have we have one two beams so that means we get six reactions this is equal to three times two three times two that equals six and this one is statically determinant so we got another one that's good and then lastly we're just gonna go here so we got uh you know one reaction there one reaction with this roller so sometimes you have rollers that connect things rather than pins but this is a you know this is this is another one and then two three four five so so this again we can see hopefully you can see one two three or five reactions five reactions which is less than uh how many beams do we have we have one beam and a second beam so we got three times two it's less than six so that's going to be also unstable so we got a couple that are we got a number of them that here that are unstable i mean with this one even um what you can see is you have kind of a uh you have a rotational instability here if you think of it like that there's a rotational instability as well so when we're all said and done we get two that are statically determinate we have two that we can we can deal with okay so so sometimes i mean it's just it's it's getting those numbers down but hopefully you can get a little bit of this where like you start thinking like okay state of the determinant and you don't have to write these things down you're saying okay one two three four five six and you can just say six and there's two beams six equals six so we're good okay are they gonna ask you all this maybe maybe not could you send a similar thing with trusses yes you could this is kind of the big picture how to look at it though okay got three questions left let's let's keep going here and this might be the hardest one of the three questions i don't know maybe it's not but oh man actually when i first saw it when i first was reading through the fe spec and i saw determinacy and stability analysis this just brought me back to consulting where we did conceal stability analysis the damn sections for a living i mean like we get paid money to do this stuff i'm not joking i'm not making it up you get paid money to check to see if a dam is stable okay i so i way oversimplify this stability analysis because i just decided don't assume any uplift okay but yeah is that a real is that real probably not but let's just go with it for this this question okay the question says we have a concrete dam section below unit weights 150 pounds per cubic foot united water is 62.4 pounds per cubic foot assume the coefficient of friction between the base so down here the base the ground is 0.35 and uplift is not a consideration so we're not i mean maybe when we get to geotech we'll deal with some of those flow nuts uplift all that fun stuff but we're not going to deal with uplift right now okay stability else is the dam section based only on the weight of concrete and lateral water pressure indicates the dam will most likely do what it'll resist sliding and resist overturning it'll do none of the above and it'll fail and fail i don't know so basically we have to check two things we have to check sliding and we have to check overturning okay is this doable i think it is let's take a look though and see so if i just redraw my damn section here for a second so ah i probably could have drawn that better what forces i mean let's think free body diagram for a second what forces do we have acting on this we have some water force we've got some concrete weight right in the concrete weight i'm going to break up into kind of two sections here i'm going to break up into you know this this maybe this this rectangular section and then the other section i'm going to do here is going to be kind of this triangular section because the weight's going to vary so if we think about this we have to think about well how much does this thing weigh how much does this thing weigh and the way that i think about it is it it's sort of kind of kind of sorta actually it's just like if we have a uniform load up here and a you know horizontal load or i'm sorry a linearly sloping load here those are kind of the forces that are holding this thing down now what's what's pushing back obviously there's gonna be some normal force there's gonna be some friction force okay so let's just see what's going on so uh let's take a look at this and see uh how how this works right so um so as we get started here we have to find a couple things we have to find this pressure down here what's that pressure do you remember do you remember it's gamma water times h remember that um what's the what's the pressure so to speak of this this concrete weight the concrete weight is just going to be the gamma concrete times its age its height right so that's going to give us these these pressure values so if we if we think of this you know this one what's that well we have 62.4 pounds per cubic foot so let's go to a one foot section what do we get we get 62.4 times what's the height of the water the height of the water is 24. so 62.4 uh let me just write this in 62.4 times uh 24 feet what's that going to equal it's going to be about 1500 1498 and 1498 what's that pounds per square foot okay and then if we do the same thing for the concrete what's the concrete the concrete's going to be 150 pounds per cubic foot times its height its height is actually a little bit different it's going to be 26 so maybe there's some gates in this dam and the the water's not going over it okay so let's look at it 150 times 62.4 is going to be 93. i'm sorry i did that wrong 150 times i wrote down 26. they said the wrong thing this is going to equal 3 900 pounds per square foot and if you think of a one foot section this becomes pounds per foot okay so nothing super crazy there and now all we're trying to do is we're trying to figure out what these forces are so we know that we're going to some resultant kind of forces here we're gonna have this horizontal force we're gonna have this the weight of the concrete one you know kind of like weighted concrete two i'll just do it like that but let's let's figure out what those forces are so h is gonna be what it's just going to be this this pressure of the water right times its height again times one half so this is going to be one half of what 14 98 i'm going to change this to pounds per foot to do a one foot section just a one foot section and multiply it by one foot just do a one foot section of the dam and then i'm going to multiply this by 24 feet its height again to get that h force so if i do that 0.5 times 14 98 i times 24 is gonna be like 17976 so almost 18 000. i'm just going to call that i'm just going to call that 18 000 and just round it off 18 000 pounds okay let's look at the weight of the concrete one what's that it's just gonna be this this three point nine kips times how far does that act over it acts over eight feet so times eight feet equals what so three point nine uh times eight is 31.2 kips okay so i'm just doing conversions on the fly here a little bit to go from pounds to kips right 1 000 pounds per kip and then this wc2 what's the weight of the kind of the triangular section so this is going to be similar it's going to be one half of 3.9 times uh it's not eight feet anymore now that the length of that triangular section is 15 feet and kips per foot i should probably put in here kips per foot uh so what do we get so one half so 0.5 times the 3.9 times 15 and i get like 29.3 roughly so 29.3 kips so these are kind of the the components of forces in that free body diagram so now let's now that we have those let's tackle this a little bit right so let's just start with sliding right so so for sliding this is kind of a statics problem i mean it's this is honestly a statics problem the stability stability analyses are kind of just like let's let's do statics but for sliding what do we what do we want we want or what we want is this thing to be stable so for it to be stable what we want is the horizontal driving forces so the horizontal forces you know the driving maybe we could just put driving forces has to be less than what the friction force at the base to resist sliding right so this has to be true to resist sliding does that make sense so the water that's trying to push this has to be less than the friction that's available so this thing resists sliding okay so what's the horizontal force the horizontal force is just as eighteen thousand pounds right so the horizontal force is easy this is just the the driving force is h equals eighteen i'm just going to convert here 18 kips we already know so the water force trying to push this thing is that less than the friction the base what's the friction of the base the friction of the base is developed how the friction the base is equal to mu times the normal force right so what's this we were told that the frictional coefficient was 0.35 in the problem above times what the sum of the vertical forces essentially sum of the force in the y direction you catch that so so just like friction and statics it's the same type of thing what do we get we get equal to 0.35 times oh what's 31 what's what do we get when we add this up 31.2 plus 29.3 is going to be 60.5 and when we do that out we get .35 times 60 uh 0.5 and i get like 21.2 kips so the force of friction is 21.2 which is indeed greater than our horizontal force so this structure therefore stable in sliding okay so it's stable and sliding it's not going to go anywhere that the friction they might be saying well is this going to is is the friction going to push it back and no only you only get as much friction as you need if if the horizontal force is bigger than 21 then there would be a problem but i've never seen a dam go travel back upstream into the into the river because friction pushed it up there right the only problem you get is when the water pushes it down right so the only problem you get is when this horizontal force is bigger okay so it is stable and sliding so that means uh we can just go and cross out the phalan sliding felon sliding okay so we know it's going to resist sliding and then let's take a look at overturning so what's overturning here so over turning you guys doing okay i mean you guys probably are sitting there with like dinner but who knows maybe maybe not but so what do we have we overturning what do we want for overturning so let's take a look at this and basically what we want here is is we want the overturning moments we want you know to to be stable we want this to be less than the resisting moments so in other words we want more holding it down than it's trying to turn it over okay so what do we get well what's what overturning moment do we have here well the only thing that's trying to turn this over right is this horizontal force the horizontal force is trying to totally push it over okay what's trying to hold it down the the cut weight of the concrete is trying to hold it down that makes sense so so basically we get two moments from the weight of the concrete and somebody's probably out there saying what about the normal force well uh technically normal force doesn't really work if the water is too big so the there isn't if the water starts to overturn too much your normal force goes sort of goes away but we'll uh we'll go from there so let's let's just look at those two sets of moments and see which one's bigger and then we know if it's going to be stable so let's look at overturning moment so what's that it's going to be the horizontal force times its moment arm so what's its moment arm it's just going to be you know one third of this distance or 24 feet what's this 24 feet over three which is going to equal what eight feet okay and this is where uplift really comes in but we're just ignoring uplift on this one because that's what the problem says so 24 18 times this this this eight feet so this is 18 kips times eight feet so if i plug that in what do i get 18 times eight is 144 did i do that right i think so i don't know let me check my answers here uh but hopefully it makes a little bit of sense yeah that's the number i got it's just a i had remembered a different number but who knows 144 hippies so that's that that looks good the resisting moments what it's going to be wc1 times its moment arm and um you know and we're going to add in wc2 times its moment arm and let's go find those moment arms and then we'll we'll figure out whether this thing's stable or not so the moment are for wc1 we're going all the way here to point a right this is point a what's that distance well to get to you know if i come to my big diagram here this thing's gonna be right at the middle of that eight feet so to get to there i need to come all the way to here so what's that 15 plus eight over two do you see it it's this whole 15 plus half the eight so 15 plus half the eight that's going to be 19 feet and then similarly we're going to have this weight of the concrete here which is just going to be two-thirds of that 15. so two-thirds of that 15 is going to be 10 feet so we have that 19 feet we have got 10 feet so this is going to be 19 feet is going to be 10 feet we can plug and chug this one out and if i put the wc-1 in i think this was like what 29 i don't remember i got to go look back all right no we got 31.2 times 19 feet uh plus what 29.3 kips times 10 feet and hopefully you can see that this is just going to be a lot bigger uh but the value i got was like 885 or something like that 885 a kit feet and this is a lot bigger than our overturning moments so that means that this is going to be therefore i'm going to say i'm stable in overturning and yeah stability analysis get more complicated especially when you start looking at uplift assumptions i i mean the the guidelines for uplift have changed you know at various points over the course of the past 50 years but it but what we have here is is an answer that we're going to resist sliding and resist overturning and we're stable okay are we going to get a question that that's that that's this crazy probably not but you know i it just brings me back to the fun of consulting so let's let's go from there so let's keep going we got two more questions these two are are kind of on the sort of quicker side but elementary statistically i'm sorry statically indeterminate structures so elementary is good we like elementary uh we like things not to be too crazy but what we see here is if we start looking at remember determinancy we start looking at this we got one two three four we got four reactions but only one beam that means it's statically indeterminate we have too many reactions for what we are trying to do so this is statically indeterminate okay so we need to go ahead and solve this one so let's go back to our reference handbook and this is where in the reference handbook we actually have uh some things that are going to help us here i i threw this one in just because i think it's it's super super useful but here we have elementary statically indeterminate structures by force method of analysis right so here what this is saying is it's saying like let's take this this this roller out at first and we have some basic beam formulas for what this looks like so five wl to the fourth over three d4 ei this is some basic beam formulas for the deflection at the middle of this beam without that roller support and then the question is what force do we need to push back up on that thing to get these two deflections to equal each other so essentially all we're doing is we're going to set this deflection uniform load without that that roller equal to this deflection you know this is the point load to push it back to zero okay so if we know those two we're going to work basically all we're doing here is we're going to equate those two and uh push them back together okay so uh that's that's what we're gonna do so let's write that equation down and solve it so if we come back here we're gonna say 5 let me get the right color here but we'll say 5 wl to the fourth over 384 ei equals pl cubed over 48 ei and this is so cool because somebody i know somebody probably because we have it we're given a steel beam here somebody probably went and looked that up in the manual so i'm gonna go find i but the good news is look at this look at this e and i both cancel out in fact not only e and i cancel out but we have three of the l's that cancel out so when we look at this what we're left with is just five wl over 384 equals p over 48. you see it so so this gives us something but what are we asking for the magnitude of the vertical sport reaction a so so we need a but this is going to give us b okay this is going to give us b so p in this case equals this is the reaction at b okay that's going to be helpful to us because we need to get there first and this is how we're going to go through this process so we're going to start with this and we're just plug in so 5 times w which is 2 kips per foot times the length the length is going to be the total length of 18 feet uh divided by 384 equals p or i'm just going to call this b y right this is this is really that reaction v y okay so b y uh over 48 and rather than doing over 40 i'm going to cross multiply by 48 and just put the 48 on this side so now i can directly solve for b y so b y is going to equal if i did this right let's look 48 times 5 times 2 times 18 divided by 34. i got 22.5 anybody else get 22.5 uh 22.5 kips so that means this this you know this force in the middle is 22.5 kips and then we've got two forces here and this is symmetric too so so what you see here is this a y and c y due to symmetry they're going to be the same and how much total force we have acting down well 2 kips per foot times 18 feet equals what so 2 times 18 gives us a total of 36 hips so now we have one thing left to do all that we really have left to do is we we just have to say okay let's let's go here and let's just sum forces in the y direction come on not wanting to click for me we can sum forces in the y direction some forces in the y direction equals zero and what do we get right so we get a y essentially a y i'm just going to say 2 a y because a y you know by symmetry a y equals c y 2 a y plus 22.5 kips minus this 36 kips equals zero so we get a y equal to what 36 minus 22.5 divided by two and i get a y equal to 6.75 kips which fortunately for us is also uh an answer here okay so i will highlight that and be happy with it so again what we got to do here is we have to we have to know kind of like where to how to deal with these elementary statistics statically indeterminate structures um we have to recognize we can't just take like a tributary area here okay um we can't just take like a tributary area uh and and solve this sorry my mouse stopped clicking this is just annoying me but um like we can't just take like oh well we're gonna take four and a half times two and that would give us nine kips which would be uh we should be wrong right so if we do uh the nine kips that's wrong right four and a half times two that's an easy answer to get but it's the wrong answer so don't so don't go down that path it's not just a tributary with you know half the distance 4.5 and and that's because the the force that this middle support does uh does get a little bit bigger based on the the the the way that this beam um acts continuously okay so let's keep going we got one question and then we will uh be done if let's see if my computer wants to respond here maybe we'll uh if not we'll we'll come back and do this one another day or i'll i'll shut it down and restart it i don't know but um let's look and see if i can make this happen we have one left and uh it's not liking me here okay um maybe okay i think we i think we're making it work okay uh there we go sorry about that let's see if we can make this work here okay so statically indeterminate structure so again we have a beam we've got a statically indeterminate structure here we've got too many reactions we've got three fixed reactions on one side three on the other side we only have one beam and we gotta we have to figure out how to solve this so again when we come to our our our problem here what we're going to do is well let's just first draw a free body diagram and see what this free body diagram looks like so if we take this thing i'm going to draw my free body diagram i got what a y b y ax bx and some of you are just tempted to stop right there because it looks easy but don't do it because it's going to cause you problems okay don't do that it's not going to work but what we can do here is we have to remember you don't forget your mother didn't i say that already um but what we have is some moments here sometimes you'll see this mab and mba depending on your sign naming convention but you have to deal with those moments so you can't just take like summer moments about point a and solve because we have these two fixed end moments that we don't know so that should ring a bell right when i say fixed end moments that should ring a bell that there are fixed end moment sheets do you remember those so if we come back and again we're back into this this the reference handbook if we come back into here uh we have some fixed end moment sheets and i can hopefully find this can i there it is we get the cool thing is like maybe the book we use is has like a whole table of them right the the reference handbook only has like two fixed end moment sheets so if you're looking for a big long table they're not going to be there this only has two so we got w l squared over 12 for uniform load with between fixed supports and we got p a b squared over l squared and p a squared b over l squared that's it and you might be thinking like that's great mark but um this asked for a vertical reaction it didn't ask for a moment in order to get the vertical reaction you have to find the moments first so uh if we take a look at these we're going to have formulas p i think this is p um p a b squared over l squared and p uh p b a squared over l squared so those are those moments and this shouldn't be super super tricky to plug in um we should be able to plug those in right away and solve so let's let's this is going to give us our moments which we'll then use to go get our vertical force so let's let's just plug in here and what do we get we get 9 times 9 kilonewtons times a which one's a okay so if we come back to our diagram here a is on the left b is on the right so a is going to be on the left this is going to be a this is going to be b so what do we have we have 5 meters times 10 meters squared divided by 15 meters the total length squared and we can just keep coming down here and solve what's 9 times 5 times 10 times 10 divided by 15 squared so i got like 20 kilonewton meters here so that's at the moment at 8b okay and then on the other side uh let's do the other side here so p nine kilonewtons times 10 meters times five meters uh squared all over 15 meters uh and we'll square that so if i plug that in 9 times 10 times 5 squared divided by 15 squared i get 10 the moment of ba is 10 kilonewton meters once you have that what are what are we finding we're finding this um the reaction at support a okay so the good news is if we have the reaction that support a we can sum moments at point b normally i like to sum moments today if you like to sum moments today go ahead some moments a day and then some forces but here i'm just going to sum moments at point b this is going to equal zero anything that's that's you know in this direction is going to be positive so i'm going to start with m a b that's positive okay this one's positive then i'm gonna have minus a y times 15 meters the whole the whole distance here and then i've got this nine kilonewton so i'm going to add in 9 kilonewtons times 10 meters i'm going kind of fast with those signs there but again rotation it's causing counterclockwise rotation which is positive and then what else the only other thing that caused the moment here is mba and i'm going to say minus mba because that's causing clockwise rotation has to equal zero we got one unknown and we're done okay so we got one unknown let's go solve it send it crush it and uh i don't know go win the gold um so what do we got we got mab which is 20 kilometers i'm gonna put this whole term kind of on the other side of the zero um just to make it easier times nine kilometers actually i'm just gonna say nine times ten is what 90 kilonewton meters and then minus this 10 kilonewton meters uh equals what a y times 15 meters so we're gonna get a y equal to what's this we got 20 it actually works out kind of nicely doesn't it 20 plus 90 minus 10 that's 100 divided by 15 is going to be what 6.7 kilonewtons so most nearly most nearly we're in between the the six and the seven but most nearly is gonna get us to seven okay so oh man elementary statically determinate structures did we have to do like moment distribution or like you know some other crazy crazy method no we didn't here but we did have to understand that that moment or the fixed end moment diagram up here right up there we had to understand that diagram we understand how to put it in how to solve it but also what to do with it once we did so man there's so much in in structural analysis i wish i could teach a whole course on actually um i do but that's you know it is what it is it's it's in structural analysis you're probably going to end up using more reference handbook resources to a certain degree than in statics right you're going to have to go look up some moments of inertia you're going to have to maybe do some more unit conversions and that sort of thing but that reminds me of a story okay this is all done actually this reminds me of a store my grandpa he's like a hundred now like literally like legitimately a hundred he sends jokes every once in a while and um and he he sent this joke so matt to answer your question yes this is totally i could totally see this question on the fe 100 like this type of thing this is an elementary statically indeterminate structure i could totally see something like this on there i don't know they give you these these are after these formulas i could totally see something like that it's not super crazy um but in case you don't have resources you might think of this gentleman um this is for my grandpa there's a gentleman who um who was sent to jail and and he was really sad because his dad had always planted a garden his dad was getting older and he normally dug the garden up for him but he couldn't because he was in jail but he still was able to use his resources because he um he sent his dad he said well his dad sent him a letter and said hey i'm feeling pretty bad because i can't dig out my garden i'm getting too old and if you were here and not in jail you could do it so his son sent him a letter he said dad dad don't dig up the garden that's where i buried the bodies love your son um well before in the next morning fbi agents and local police arrived and dug up the entire garden without finding any bodies um they apologize and it's the old man and left um the same day the old man received another letter from his son it read dear dad go ahead and plant tomorrow um the garden should be dug up now that's the best i can do under the circumstances i gotta love my grandpa you know what that guy was resourceful even though he's in prison he uh he made use of it um honestly you guys are resourceful i know you can do this this this type of question isn't too crazy it's something you can make work but oh man i i appreciate you tuning in here and going through some of these problems with me and uh hopefully if nothing else the joke is worth it you know i'll try to come up with a good joke maybe it's not good but it's uh it should be fun so hey um if you guys have questions or you're in my actual class here feel free to stop by tomorrow i've got some time middle of the day so i'd be happy to to chat with you but um you know we can uh we can go from there so next week we'll be doing structural design we'll get into some of the concrete steel design and then just keep going so hey thanks guys for tuning in i know isn't it crazy it's something so stupid it could be funny but uh enjoy enjoy enjoy thanks for tuning in with me and i'll see you next week so until next time guys keep working hard moving onward and upward you