 All right, any questions homework day is it yeah, all right, let's go, okay All right, well on the way out. I don't need them now. I'm not gonna grade them now Now we today we really bring together Some of the stuff some of the stuff we did a lot in statics in fact it's important enough what we did statics We're going to spend a couple days reviewing here We're going to start looking at bending most of it's going to be based around simple beam design and We spent a lot of time last fall looking at beams in some kind of transverse loading where it might be Uniform loading might be point loading of some kind We even had cases where there was some kind of moment required as it would have been the case In some kind of cantilevered beam like that All of those type of things well, then there there might even be the type of thing where we had some kind of load that would contribute in a transverse fashion In that there's a Horizontal load there, but now we also Might have axial loading We didn't really look at that kind of loading much if at all and in Static slash fall well, we did we did have axial loading when we had two force members or two force members Could only be But we didn't have to often put on a tire a tire beam itself in some kind of axial loading But we we can and will hear because we certainly looked at the stresses call Caused by that type of loading in fact, that's how we opened this course the very first thing We looked at the very first day. I think was axial loading They contributed to normal stress So we're going to look at this kind of thing a lot We'll we'll open with Just the normal type of things we did before remember we had Simple supports that we could statically Solve for and then we put some kind of loading on there We'll continue to do that type of thing now But as we've seen already we can also look at statically indeterminate type loading the simplest being a Beam that is pinned in both supports rather than the usual We had last fall where we had a pin and a roller support because we couldn't solve statically for any type of Loading that we might have had that would require more than three reactions on a two-dimensional problem in Static we only had Three equations that we could use on our two-dimensional problems the three equations being of course some of the forces in the x in the y-direction zero and Then just some of the moments in the plane. We are looking at we're zero so we're going to have to add to that as Need be as we start to look at the real reaction of these beams We'll very quickly here get to the point that these beams Will undergo some kind of deformation This one might do something like that because of the the loading We're going to use that Kind of thing just like we did for our statically indeterminate beams Actually this one would probably go up a little bit down a little bit Something like that. We're going to actually be able to calculate these these deformations from the neutral position at various places And use that kind of thing to solve statically indeterminate problems that we put before a Little bit before we get to that. We are going to look at what kind of cross-sectional shapes we can use whether they're i-beams or built-up beams, which are the type of things you might do if you're using a Lumber to put some pieces together All kinds of different things or possibilities that we'll be able to look at we're also going to look at what happens At these joints that we have to either nail and or glue together where these Beams are attached to each other None of that kind of stuff did we look at in statics. We didn't look at the cross-sectional shapes We didn't look at the cross-sectional areas We didn't look at the deflection of any of these beams and what these shapes and areas have to do to the resistance of these Materials to severe deflection some deflection is Just one of the facts of life like friction is but we don't have to undergo severe deflection So even to the point where you could consider this to be failure of the beam All right, so that's where we're going now To get us there first we're gonna have to do a little intense review In fact, I think it's important enough. I believe I've Set aside two days for this as we're going to revisit Shear and moment diagrams you guys Bobby you look forward to that like those You guys seem to have it pretty good last fall I don't remember the feeling going through the static class that a group had this as good as you guys seem to but We got some newbies here. They're barely away Frankie, okay So we're gonna we're gonna revisit that It's an application of only our Static equations, but it's going to lead us to the ability to design the beam cross sections and look at The influence of these forces on the material themselves So to start with we'll we'll just do a little problem here So we've got a beam that's got a pin support there Roller support there and it has two loads On a two-point loads there 40 kilonewtons there these are typically gravitational loads something set on that beam whether it's a wall That's holding a floor up above and that's why it looks like a point load All kinds of things for possibilities So two point five meters of that and overhangs Then we have three meters from that support to the load and then another two meters out to the end So Remember what we're trying to do with the shear moment diagrams is look at the internal forces In the material itself the shear moment diagrams is a graphical display of that and it allows us to Determine what the internal forces are so that very shortly here Next week we'll get to it We'll start designing the cross sections that can withstand those material those internal forces with particular Loads of some kind So if you remember One of the first things we need to do is determine what the reactions are This is a statically Determinant beam so we can determine Using our statics equations the reactions of those two supports So do that real quick is kind of a warm-up for both the day and the semesters We haven't done too many of those kind of things in a while, but we've got to have those right otherwise, we don't have the Internal forces right to come in right after this the two reaction forces Might not be up, but it certainly looks like we'd expect that to be the case We don't know how much there's going to be nothing more than two Add together the equal 60, but you got to figure out how much of each one goes on each reaction almost about some point Determine what's the NDR like work. We were doing somewhere in there There's the factors of safety of two or three anyway, so something goes wrong right here Forward more uniform because I don't think they Don't agree to do me and they don't agree with each other. I don't think problem. You got that much, right? Then what'd you do? Moment any clockwise moments equal any counterclockwise moments because they all sum to zero So about be counterclockwise 20 times 2.5 D is also a clockwise or counterclockwise moment But we don't know how big D is and that's a distance 5 meters I already got your mistake And then 40 times 3 is the only clockwise moment all the counterclockwise moments Pat, where are you? Do you agree on anybody? Were you Do I need to call security or not health and safety? I'm sure it's DJ caught your mistake. Yes Jake got your mistake You agreed with them. Do you does he he didn't check did he? So you're lying to me in two things one day using my son only does one a day All right. We got B was what Jake 14. Is that right? We need to draw the free the sheer moment background. I Encourage you do them straight up and down at each other what we're going to need most is what is the maximum sheer and What is the maximum moment and where they occur? Because we're going to start designing beings to withstand those internal forces of Shearing moment selectively build up those beams So we need to know where the moment and where these Where the beams are so typically let's review what these internal forces are We imagine a cut through the material to expose those internal forces And we need to do this anywhere in a section things haven't changed much so we between a and B Nothing really changes so we can put a cut anywhere in there some position X and that will expose the sheer That's going to be needed to withstand any forces on that section and since those are separated by some distance X There's a moment there. We need to have some kind of other internal moment to resist that and we need to find out If you remember we have a convention for our signs any sheer on the right side That's up and any moment That's counterclockwise We call positive Sorry, I got it wrong with this year Down on that side up on the other expose So that's positive here That's positive moment any moment on the other side That's up Moment in the other way. That's positive. That's our convention For what we call positive and what we call negative so we can see that the Shear needs to balance that force. It's just a sum of the forces in the y direction That's going to be negative for the convention we've chosen So along some distance X And units of kilonewtons anywhere between a and b the The shear is a constant and in this case a minus 20 and we know that that's all the way out to All the way up to point B The moment itself is a factor of this X that comes from it the internal Strength of the material itself to try to resist that moment that's increasing as We move along the beam because that X itself increases This very end X is zero it increases from there. It's a negative moment in our convention It's linear so we know the moments be something like that To a maximum at point B at least a maximum for what we've done so far When we do the 20 Kilonewtons at the distance 2.5 So so far our maximum Is minus 50 kilonewtons meters That takes us out to B Where we have the reaction applied and then nothing else that happens beyond B So we can go beyond there somewhere To put our cut and now we're again at some Some distance X Beyond the 2.5 Because we need to go past B. We've already done up to B and before we get to C Obviously, we've got to have some Shear down to resist the Extra amount going up because of the support there and it's got to be of a 26 Kilonewtons, otherwise we wouldn't have a simple force balance on that section remember every portion of everything we look at is in static equilibrium and then we need some kind of Some kind of moment here. Not sure Which direction it was a go until we calculate it It's got to be enough there that it resists the moment of these two parts We can calculate the moment it needs to resist It's got to be 46 Kilonewtons That far away from that end. There's that moment caused by the support that distance is X Minus 2.5, then it also needs to resist a moment in the opposite direction a Distance X away However, this is all Negative convention So we really should have a negative sign in there Convention on that that'll do it for us 46 times 1.05 plus Is a plus 1 15 is that right minus X a minus 46 X plus a 20 X is a minus 26 X No minus 14 then minus 46 plus 20 26 oh no, I had Because we know that the moment must be continuous since there's no applied moment So let's check this at X equals 2.5 We should get the same moment we left off here, which is minus 50 do we? So 115 minus 26 times 2.5 Do that real quick If we don't get the same minus 50 then we have a mistake it is okay, so We can use that as a check Simply because we know it must be continuous from here so This is positive minus 46 this comes out to be positive 50 so we have a minus sign error Through there and it didn't work 2.5 the moment arm is this distance Which is X minus 2.5? Oh, we have it drawn in the right direction So so the moment is going to become like so Should be 46 X minus 20 X All right, so that gives us a minus 50 and That's what we wanted one minus 50 plus. We knew we wanted a positive slope 2.5, which takes us to see and we have a linear slope in between there so 5.5 times 26 minus 115 is 28 right it's 50 plus 28 Somewhere in there we can do it from either end now We've exposed a little bit of the beam past C between D 60 down 46 up. We got 14 up That's a negative convention. Oh, we didn't draw in this year here There we had which was a plus 26 and a minus 14 that matches with the amount of Support that comes up brings us to zero at the end of the beam and when you'll see it's linear and We know there's no moment at the end because it's just a simple roller and there's no way to support any moment And it's got to come into the zero at the free end of the beam Is that kind of support offers no moment support so we must finish at zero at that end of the beam They was at the rule with the no moment of the reaction is it just rollers Yeah, remember remember we had a table in the statics book to show what kind of support and what kind of reactions each support Can give in the roller support and the pin support can offer no moment So that's why there was no change in the moment here at that reaction And there since this is at the end of the beam must finish at zero moment there Some reason I just left that being had a way to go to zero No, it doesn't say there is no moment in the beam there there is bending in the beam there However, it doesn't contribute any moment, which is why there's no change in the moment here There's a change in the slope But not a change in the moment What we're looking at what we'll be able to see soon is any place the beam would have curvature There's going to be a moment in the material so you can imagine that if it was loaded like this It would curve Something like that probably anywhere. There's a curve in the beam. There's internal moment So you can see the beam curves at B It's pretty straight right here where it goes from one curvature to the other and that's about where we have them at the zero moment So now we would in fact know Right there the slope of the beam is Essentially linear as we go from one curvature in the beam to another curvature That's the kind of thing we'll be able to look at now is what what is the the Beams reaction now itself to the curvature in these beams Don't forget that all of these three diagrams must agree with each other In terms of what the loads are and what they cause What do what is the relationship between the area and here? These areas these areas have to do with each other, but no these areas are not the same Hang on I got two years you take one you take one go Well, what is the relationship between the areas and in fact there's a relationship between the slopes as well What is the relationship here remember? the slope of The shear diagram or the value of the shear diagram is the slope of the moment diagram remember that V equals DM DX if we take that and make it V DX equals Dm and then to integrate from Any one place X1 that any other place X2 and we integrate from M1 To M2 where M1 is the moment at X1 and two is the moment at X2 This side is the area Under the Vx diagram Which is That little square there for this example, so it's minus 20 Times 2.5 is a minus 50 is the area It's units are killing newtons times meters So it has units of moment That is equal to this other side, which is The change in moment between X1 and X2 and that's exactly what we got the change in moment Between X1 and X2 is equal to that area, which is minus 50 so we have those two Relationships between these two graphs that the value of the shear at any point is equal to the slope of the moment diagram The slope was minus 20 here plus 50 there minus Whatever that was 14. I think yeah minus 14 there The shear was constant in those sections, so the slope is constant in those sections And that's what we were Getting right there. That's the slope of it. That's an equation of a straight line and then also the area of the Vx diagram Equals the change in the moment between the same two spots that you use for the area the area of this middle section What was that value? 26 So 26 times 3 78 so the change in the moment is minus 50 plus 28 is 78 And so all of these things I agree so remember those two very important Relationships to always check this stuff And then if you remember we even have another one, so we'll get to that right now with another beam which we can start Right now come up with the free robot or the shear moment diagram So we have a cantilever beam That's one where one end is embedded in the wall that means there is moment support as well as Vertical support there middle of the beam got a uniform load there as if Some material of some kind of stacked there or maybe you expect it even a human Load to be there and then so that's That's eight feet from the end of the beam to that point three feet. We got some kind of Arm there of some kind that's supporting its own load of ten tips sometimes, so maybe there's some kind of Motor running something there with this Arm there Two feet I got that picture complex loading remember you can do it from either end But the relations are all the same. It doesn't matter We're still looking for the internal Resistance of the material to this loading. Oh, I need to give you the Load distribution here three kips about this load here on this bracket arm What do we do about that on that bracket arm? What does that mean to us in terms of replace that with an equivalent load on the beam as This ten kips pushes down that's going to cause this arm to push down As well with ten kip that vertical load because of this arm is going to be transferred over to that point that point C however Because of this moment arm. There's also going to be a moment Exerted around that point in that direction an applied moment then of Twenty-two feet so we can replace that bracket arm with this kind of load at point C Nice when you already know the answer Where to put the Axis I know that it needs to be high I've already done this so I'll save you some erasing We're going to have nothing but negative shear and negative moment on this problem We don't know that normally going in But I happen to know it so I'll save you a little trouble when you're driving your graph so your notes aren't so messy So we'll make a cut somewhere between a and b X out from the end We need to do it our cut somewhere between a and b. That's that's where We that's how far we can go before things change. We have a uniform load till there After that the uniform load stops, so we'll have to redo it after being oh You need to have found the reactions. I hope we should do that We'll put it in real quick. We can double-check it supplies a moment 318 of some amount negative just the way We can tell from what we've already got loaded How much shear must we have at this Little exposed piece right here Remember it's a it's an artificial cut. We've done there. We're looking for the internal forces We have a uniform load of three kips per foot for x feet So that'll be three kip and it's negative in there. Oh, sorry three Negative our convention says that we make it negative at x equals zero Minus three up on that end is our negative convention goes to a maximum of this 24 Yes, how do we figure out the moment a single point load then of? 3x since there's no other vertical loads of course that's equal to the shear and so the moment is That load times that moment arm and it's a distance X over two away the convention. It's minus five point x squared No, it's something like that So it looks like it goes asymptotically to that it doesn't necessarily and at point B Where x equals 8 feet? The moment is 1.5 times 8 squared Which is? What how much? Negative 96 that change in moment should be the same as the area Under here, which is one half times the base eight times the height 24 Is 96 so that made sense again the area here between the same two sections. All right, we'll We'll finish that up Monday, and we'll we'll do some more because we need to get these right Again what we're looking for is the maximum shear the maximum moment and where they occur So we can start designing the cross-section of the beams and Choose the material to withstand those loads at those internal points