 Alright, we're going to change a little bit now from the type of loading we've done so far. Well, we've done two types of loading, axial loading where the forces are in the direction of the greatest dimension of the, so something like that. We also have looked at torsion, which is some kind of twisting load on the piece as we might see in driveshafts and the like. Now we're going to add to it the, actually the main type of loading that we saw in statics, which is the loading that causes bending, which is for most of what we'll do, at least certainly in this early part, transverse loads, transverse to the, across the axial length of it, across the greatest dimension. And that's, if you remember, an awful lot of what we did in statics, a whole bunch of those kinds of names, spent some time learning just how to find the reactions. What we're going to review again then, this morning, is the shear and moment diagrams we get from that, because what's going to be a big concern to us now is how these beams respond to this kind of loading. It's not any great shakes to know that these are going to bend, something like that. But any place you've got a curve in the beam like that, that's internal moment, we need to look at that, we need to see, can the material withstand that, those kind of loads, those also of course shear across here as well as the internal moments. And then a little bit later, more near the end of the term, we'll actually look at what is the displacement of the beam at any one place, its actual deformation, just like we've done with the other two loadings, we looked at the loading first, made sure the material could withstand the loading, and then actually looked at the deformation. And if you remember, for both of these, and it'll work again for this, it allows us to determine the reactions of statically indeterminate type beams. And if you remember, our simplest type of beam support was a support of some kind like that, such that it was pinned at one end, but we had that little roller at the other end. Because if we had actually pinned it at both ends, which is pretty much what any of us would have done if we were actually building something, if you're going to build, put up roof choices in a garage, you're going to nail them down at both ends. You're not going to nail it down at one end and then put some kind of roller at the other end, nobody built the garage down. But this type of support is statically indeterminate, meaning with our equations from mathematics, which was nothing more than the sum of the forces and the sum of the moments, with those three equations, we could not solve for the reactions. If you remember, pinned type supports, and this isn't loaded yet, but if it was, we know what to do with it. The pinned type supports supply a reaction in both of our ordinal directions, and that gives us four unknowns, and we only had three equations. So it's going to be this actual deformation that will allow us to find a fourth equation where we're going to actually come up with ways using the deflection of the beam as a way to solve the statically indeterminate problems like we have there. Other ones we looked at, we had, we were certainly able to solve the reactions for that kind of beam when loaded. However, if it was that kind of support, then that we couldn't solve. But now we're going to be able to, as we add in the actual deformation of the pieces, just like we did with axial loading and torsional loading. So let's do a couple problems. We'll spend today just reviewing our ability to put all this together in terms of a shear moment diagram. Well, one of the things we're looking for the most is not necessarily the internal shear in the moment at any one place, but where it's a maximum because we could certainly come up with the ability to design beams that can withstand the maximum moment, the maximum shear at a certain place. Maybe we don't need quite so much beams somewhere else. And that would allow us to save some of these beam designs that we're going to be working on now. A little bit of a reminder of what we're going to be doing, of course, is graphing the internal loads, the internal shear and moment. And if you remember, as we did that, we had a convention for what's positive and what's negative as we look at these. Any shear that's down on the right side and up on the left, and by that I mean as we expose the internal structure of the beam with our imaginary cuts through the beam, of course these beams are imaginary anyway, but we imagine a cut through those to expose those loads. And then any moment that is in that direction on the two either ends, these are what we arbitrarily call as positive values. Any time we have actual shear and moment in that direction on either end of those exposed cuts, we'll arbitrarily call those positive. What we're most concerned with is where the maximums are, but as you'll see with the, when we really get to the design of these beams, the direction is going to be important, too, because not all our beams are going to be symmetric in all directions. For example, if we're looking at a simple cross beam like this with some internal moment, we're going to find it's quite different which direction the bending moment occurs, whether it's positive or negative, for a t-beam like that, that's the cross section of the beam itself. Everybody's pretty familiar with i-beams and of course just regular beams, but we're going to look at all of the, lots of the possibilities, we're going to have to look at all of the possibilities, but we're going to take into account all of the generalities that we can as we start to design these beams. So in a few weeks, actually probably just right after break, if I remember, right after break we're going to be right at looking at what the cross sections are and what those mean and why i-beams and h-beams and box beams and other kinds of cross sectional shapes are so useful and important to us. So let's do a problem just for review, just like we would have done in statics a couple weeks ago just before Christmas break we were doing stuff like this. So at some place here we'll put a pin. Now remember, all that does is keep the beam from either going up and down or left and right, but it doesn't keep it from turning. That's a simple pin support. And then we'll put a roller score right at that end. And then the loading that we'll put with this is 20 kilonewton load there and then not quite in between a little bit to one side between the two supports 40 kilonewton load there and the distances between each of these spots 2.5 meters there and these are of course in all of these are in meters 3 meters there and 2 meters there and not quite the scale, they're pretty close in all of these in meters. Alright so there's our general setup just for help we'll label the main spots the places where something changes in the loading either a support comes into the problem or some kind of extra load is added which is your part of the load has changed. So if you remember what we typically did was start from one end, measure x from there so that we'll be able to get the shear and the moment the internal shear and moments as functions of x and then we'll be able to graph that with these and I thought about it since we last did it a little bit and I came up with a way that might help to make sure that we get the signs right on these as we go to graph these one thing that can work is once you put in the unknown internal loads that we've exposed with our imaginary cut, sketch them in as always in the positive direction then just simply solve for them and graph the values that you get straight away from that for example it becomes a simple matter of solving the forces in the x direction they're both in the same direction as drawn so we'd say 20 kilonewtons plus v equals zero then when we solve for v it's got a negative sign on it which is useful because that's the number we have to graph again my recommendation is that you draw these above and below each other as we go through these so that everything just makes sense you can use certain places to double check so we'll graph the shear as a function of x as we go and we already know between A and B we don't have to do anything else between A and B because none of the load changes in that region but from A to B we know that the shear is minus 20 and since we put in our drawing the positive direction whatever value comes out that's exactly the value that we need to graph so the minus is already on there we don't have to reinterpret anything it's a little bit different than we did last fall last fall I think we typically looked at a down load here that we knew the shear would be up we drew it that way and then had to interpret it for the drawing if we put it in positive just sketch it in positive automatically every time then solve for it and now our sign is there so that may be easier maybe you prefer not to do that it's up to you but that's what I'll try to do from here on out and then when you take statics again in the fall that's what we'll do there oh by the way did you see the note I put on that Binghamton has some online courses for the summer I believe that may be of use to some of you statics is one of them but I don't think if I remember any of you need that however there are some other courses that you might want to take online there just to put a little bit more on the bank at state prices without having to pay residence you might have to move down there is anybody thinking of going to Binghamton anybody thinking of going to Clarkson by the way thinking of it we have several scholarships available what I need from you is a paragraph of why you might deserve one and I can help you put that together that's also true for RPI coming up in just a couple weeks so start thinking if you want to go to either of those schools about putting a paragraph together for us so that we can and there are substantial stocks about ten thousand dollars a year alright so again and then we sum the moments and know that they'll sum to zero we have a clockwise moment between the v and the the two v's a distance x apart we know that's in the negative direction and then that's got to equal the also a clockwise moment and that's our positive direction for these that's not going to work these are in opposite directions these are in opposite directions so yeah so I think we're okay alright so v we know to be minus twenty so when we put in the minus twenty we get two minuses when we move it over to the other side we get m equals minus twenty x Newton meters and then that's if you remember what we graph on the moment diagram Newton's kilometers starts at zero because this is just a free end a free end never has any moment in it internally so starts at zero and then goes down at a slope of negative twenty to a value of x is 2.5 2.5 times twenty is fifty and so we get that kind of picture down to there helps to hatch it in a little bit I think just to illustrate what's negative what's positive as we go through these alright we can continue on nothing again changes until v once we get from v to c we do have a bit of a change so we draw out to there we've got that load there twenty kilonewtons we now have one of the supports at 2.5 one of the reactions that's a reaction at b oh we didn't do those so we have to figure out what those are but that's a fairly quick manner I'll put them in for you you can check them b happens to be forty six kilonewtons remember we get that just like we were doing the first month or two in statics do that on the whole beam I was just so intent on getting to the internal loads I wasn't paying attention we had to come up with the external loads and when we get to it d is fourteen alright so we need to figure out what the internal loads are draw them in as positive every time then when we solve for them we'll get the direction we'll get the number we have to plot straight away so summing for the shear there, the internal shear we've got twenty down forty six up and let's be not as used to doing it this way so I've got to double check what we get then we'll start with the v we know that's positive v let's see the twenty is in the same direction the forty six is in the opposite direction that's all for no news let's see if that works out maybe it's better the way we did it before but what we did before is everything up and everything down so that's what we got here then so v equals then minus forty six but it goes over there plus twenty six which remember is what we'd expect we have a reaction of forty six we know there's going to be a jump of forty six that puts us at the plus twenty six goes all the way across to C because there's no change in the load there so that jump from down here to up there is in the same direction as the reaction and so it works out then we sum the moments remember x from from the x as measured from the left side so we know we've got forty six x kilonewton meter that's the moment caused by the first one if we're doing moments about that in A then minus v x plus m equals zero or m is that internal three moment there oh sorry this is not forty six x it's forty six we know that distance it's two point five all made a review on this if you remember we've got lots of double checks and troubles here and there anyway so solving then for m forty six times two point five is one fifteen the the v that we had in there is the twenty six so yeah so we're okay twenty six x minus one fifteen kilonewton meters so we have to graph that it's got positive slope because this is twenty six x the intercept is minus one fifteen remember that's where x equals zero so that's somewhere back here I don't know if that's that much useful to use but what we can do is notice that this is just a simple pin support which means there's no additional moment caused by the sport so our line starts from the point where it left off we can then draw our line from there with the slope of twenty six from there and over a distance of three then we can find out how much that moves up it goes up to about twenty eight so if that's fifty we know that if x equals um three beyond point b we're going to go up to a twenty eight and then we can draw that straight line there and that will have our slope of twenty six automatically so again what we're looking for is for the most part is where's the greatest shear where's the greatest moment and we'll be a little bit concerned later on more concerned later on just what the direction of those moments are because that's going to tell us where the bending what the bending direction is from c to d remember we have to do this every time there's a chain in the load in some way so we have that there we have b here which is forty six kilonewtons up we now have a forty kilonewtons down at c and now our cut is somewhere between c and d on the shear in the moment and just solve for them down moments v plus twenty plus forty equal all the up moments forty six kilonewtons and so we solve for v there let's see that's sixty goes over becomes minus what minus fourteen I think which it should be because we finish with a support of fourteen over there so now we can draw in the moment and just using the value that came right from assuming positive direction gives us an automatic negative sign and now we've got the shear diagram completed and it agrees because we jump up fourteen to the end and there's no additional shear at the end so we finish just as we should let me just erase that so we've got a positive forty six times two point five which we know to be the one fifteen and the opposite direction is forty times two point five plus three five point five one hundred meters and then also in that negative direction v times x just and then plus n and so all those numbers come out to be I think if we put them all together let's see the v is minus minus fourteen so when we solve for it all I believe we get minus fourteen x plus one oh five minus slope makes sense because we know we have to come to this roller support end which will have no internal moment left in the beam so we know we have to finish in zero so the minus slope makes sense and if we put the values in we don't start or we start exactly where we left off at the moment and then we're able to finish our our moment diagram the slope is v at any spot in the same place so it's twenty six there minus fourteen so the slope of the moment diagram is the same as the shear remember to the area of the shear diagram between any two spots is the change in the moment in those two spots so all of these things should check even if you do get a minus sign wrong somewhere you should be able to catch it with these checks that we can do and so let's even use that to our advantage and do the other one cantilever beam which means it's embedded in the wall point there's a uniform load distribution that's in units of pounds per linear distance of the beam itself so three kips foot there out to a distance of eight feet then just to just to be nasty we've got an extra structure in here upon which there's a load put it up kind of high because I want to get the drawings in below it ten kips and the distances are this is three feet is two feet there's three feet left over to the wall so first thing we can do is figure out what the reactions are so there's that load what can we do with the load caused by this little bit of superstructure there can do with that because of this moment arm here that superstructure is going to cause some bending some moment to be applied around that point so we'll replace it with that an applied moment of ten kips times two feet twenty kip feet but that ten word the ten kips down is also pushing on that point so we need to add in that effect so we can replace the superstructure there with that ten kip load with a moment and a just a straight load of its own and then also need to figure out what kind of support reaction is there C and D so that's the load of D but it also affords some moment to it as well and we can figure out what all those are remember replace this distributed load with an equivalent point load at its excuse me at its geometric center it's a centroid of area remember the magnitude of this is the area under the load curve so it's three kips per foot eight feet so that's an equivalent of twenty four kips that's only for the purposes of finding the reaction once you do that solve for the reactions you should get that this is three hundred and eighteen kip feet in that direction and the reaction of D is 34 kips you can double check those draw our shear moment diagrams and speed it up a little bit if we remember some of our graphical relations between these I have it all off hand but we only have negative shear on this one so I can go ahead and draw the axes like that the easier for us this end A chalk desk getting it it's a free end there's no load until we actually get into the beam some distance so we do know we started zero then do you remember what the shear is in relation to the distributed load do you remember what that deal is the relationship between the distributed load which we call W the reaction of X remember what the relationship is between W and V the opposite of the shear load is the slope of the shear diagram so that's considered a positive load because that's the typical way loads are so we have a negative slope of three kips per foot and that would go for eight feet so it would take us down to about 24 so I can leave enough room something like that and we can do that remembering this relationship between the shear and the distributed load without having to actually make the internal the imaginary cuts and figuring out the internal loads there also it turns out ahead of time the moment is negative so I'll draw it like this save us a little bit of trouble Kip feet you know the moment at A since that's a free end is also zero what's the relationship between the shear and the moment diagram well it's the same thing we've got here we know that the slope of the moment diagram is a negative shear so we have an increasingly negative slope on the moment which is a parabola concave downwards just where it goes to is a little bit difficult to figure out unless we remember the opposite of this is that the change in moment between any two spots is the area under the shear curve between those same two spots so the area under the shear curve will tell us where the last point is and then we can sketch in a parabola once we have that one half the base which is four times 24 what's four times 96 so we know we go down to a minus 96 I leave myself enough room we're going to have to go a little bit deeper later and then we can draw it in a parabola like that easily once we have those two points we know it's at zero we know the slope is at zero as it comes in because the slope goes down to zero as it goes down to zero coming into that end B onward to C there's no change in the load so there's no change in the shear there's nothing more we need to do we know it's going to do just that we know that the moment now has a slope of minus 24 from there on and the slope is continuous so it actually we continue the line straight where the slope left off once we're going to have to redo this drawn just because we're kind of running out of space here at the bottom of the board we know it ends at whatever this area is right there which is 24 times 3 which is what 72 I think so it's an additional 72 beyond the 96 which is 168 and that, yeah, that agrees so we're down here to a 168 minus 68 so that's our change in slope sorry change in moment from the point where we just left off to the point where we're now finishing because that was the area of the shear curve right there above it and we can continue now we know that we have to get low down so we just simply draw that on the shear curve it takes us down to about a minus 34 and we know there's nothing left over till we hit the wall where we come up 34 to the wall and finish the problem and so everything matches now in the moment diagram right at this spot we have an applied moment at any kip so that's going to cause us to jump up by 20 that takes us to 148 right about there run out of space so I just have to take my curve a little less deep minus 96 this is minus 168 we jump up by 20 that's a point moment applied to the beam and the response of the moment curve to that then we know it continues from there on with a slope of minus 34 so it's going to be negative in some measure and we can figure out just how much by calculating that area minus 34 times 5 is 170 so we go down a further 170 from where we are there then we're at minus 148 so 170 plus 48 is 318 and that's where we expect to finish because we have this applied load applied moment due to this reaction support there so it takes us down to minus 318 which is just what we'd expect to get back up to the end of the beam and that slope is minus 34 and we've got the shear moment diagram and as happens sometimes you need to re-sketch them a little bit okay so remember those relationships it didn't happen to use it but it can be very useful true that the change in the shear is going to be equal to the opposite of the area under the load curve so we have those four relationships plus any of the point values that we have and understand especially at the free ends or at the reactions combine all that with the known jumps because of either applied loads applied moments whichever might work and we can complete these what's going to be of concern to us for the most part is what are the maximum values on any of these because that's where the beams most likely to fail and we're going to start designing the beams now to prevent failure by designing the beams the size and shape of the cross-sections of the beam the cross-section of the beam is going to be able to resist and the material of course resist these maximum shears and maximum model alright so one for you to do again a cantilever beam split into thirds with a 19 kip load sorry not kips, kilonewtons three meters there and again three meters we're going to have a uniform distribution of a load so beam split into thirds like that and the last sorry not three not that it really matters we're making it up anyway but at least we'll agree with mine and the uniform distributed load three kilonewtons per meter alright so you come up with the shear moment diagram that's it, don't worry if you do well with that I have another so if you come up with the shear moment diagrams I'll help you a little bit just let you know where the axes are sometimes we need a lot of negative axis, sometimes a lot of positive some fairly even these ones are fairly even don't forget first you need to find the reactions and since it's a cantilever there's both a moment and a shear applied there just to help you out that's the direction of the moment on that one which is a little different because that 19 kilonewtons load is up so it actually makes the moment in that direction supplied by the wall so use any of the graphical methods use any point values that you happen to know just by observing the beam and if ever need be always make the internal cuts and figure it out like we originally did before we established those graphical methods whatever combination of the all of these things all of the hints that are there for the values you can make use of as you wish look at the beam and usually get at least a few of the points you might not know what the line does in between but you can place a few of the points usually at the ends and check the moment curve the value of the moment curve actually tells us how the beam is going to deflect how it curves because of the loads of course the solutions are online so you can check any so you gotta figure out the the reactions shear is usually pretty easy to come up with a little more subtle but if you just make all the checks and balances all these different clues that are here you can usually put it together because then something conflicting watched you're being careful a lot of you are missing something replace this with an equivalent point load of 6 kilodooters but that's only for the purposes of finding the reactions once you've found the reactions you're done with that you're done with that equivalent point load there so we have 6 kilodooters down 19 kilodooters up so A is actually in the wrong direction there you're going to get the A of minus 13 so that's your starting point giving you enough room that's actually the biggest value we're going to get for 19 kilodooter load the the wall is actually holding that end down not holding it up which is usually what we see for a cantilever support at the wall then you know there's no more load on the beam between A and B so it just goes straight over grabs that help a little bit shear diagrams usually a lot easier because the jumps and the shear diagram the directions of the point loads which helps a lot it's a more severe getting that it jumps up 19 to a plus 6 and then nothing happens in between shear we're not so much concerned about the direction of the shear when we start designing the cross sections and the materials and the like we're concerned with the direction of the moment so as we'll see the slope of the shear diagram is minus the moment and we get to a free end and we know we finish at zero so it's a straight line coming in for the moment diagram it helps a lot I think to remember that relation that you have change in the moment for any section is the area under the shear curve between those same two points I think that can help a lot we don't start with zero moment because we have the cantilever support there we can finish with zero moment because it's a free end but a cantilever support will supply internal moment what that means is right at that point the beam is already going to be curved some final shape of the beam will be more the slope will be zero there but it will immediately start curving and because of this unfamiliar 19 kilo Newton load up it actually curves up there as we'll see in a little bit 20 stacks actually look at the deflection got it Chris you think no applied moments so there's no jumps in the moment curve there's breaks in the slope discontinuities in the slope but not any discontinuities in the curve the line itself there's no applied moments what do you have for the moment supplied by the wall, the reaction supplied by the wall you're going to need that that's going to give us the starting point on the moment diagram it's not zero there because of the cantilever support it immediately puts a moment in the in the beam and you can figure it out by summing the moments about A and using equivalent point load of the distributed load be careful don't forget that's not actually there because of this right here in our prediction that distributed load down is considered a positive load so we put a negative on it and then the slopes match that's the only one of those relationships that has a negative sign in it well I guess it's equivalent one and that would all match with our convention of positive directions we need that moment at the wall that's in the proper direction do you have the magnitude of the magnet of MA I drew it in the right direction for you I didn't happen to with A itself but I did with MA for you so if we draw that in MA in the proper direction nobody's yet given me the magnitude we have A is actually in that direction without that wall that end of the beam would go up that's why it's negative there that's our negative direction but the moment must also be in that direction and so when X is zero we're right at the starting point there's no moment due to the shear we know that we start at MA in a must be in positive direction does anybody have the magnitude for this 9.7 that's quite what I got 8 is what I have in that proper direction so we actually start at 8 this upward 19 kilo Newton load actually makes these reactions on this cantilever opposite of what you'd normally be used to we're normally used to everything on the beam pulling it down but that actually gives it a net effect of going up that's what you're it's wrong but it gets extra credit for creativity it's an awesome try so if you know it starts at plus 8 you know what it does from there it goes down from there straight line because of the constant negative shear and we'll finish at a distance delta M below 8 which is what 13 times 226 so it finishes at minus 18 is that right so that's plus 8 so about 2.5 times that we know that it's a straight line down to there to minus 18 because that's what the delta M was you know I have to actually figure out the slope you don't actually have to graph the line you know what's going to do that the shear diagram it's positive so shear diagram then turns positive and constant so we have a positive and constant slope and we can figure out the delta M for that space that's just a 12 so it goes up to what a minus 6 what you put it right about there at minus 6 yep those two slopes minus 13 plus 6 and you finish at a minus 6 there then it should be easy to finish you're going to stay with that one Chris I may have to put that up on the projector to share with the class a rather unique answer Phil don't you wish you could come up with that answer rather whatever you have pretty awesome so we didn't have to actually do any imaginary cuts didn't have to break any of them in there and then you know how it finishes you know slope is continuous there's no break in the slope here no break in the M the curve itself there's no applied load and we know we finish at a 0 so continue the slope but then start curving it in and bring it in then to the final final little bit you can bring it into 0 and you've got the whole curve and all the stuff agrees there the area of that should be minus 6 and indeed plus 6 because that's the change in M that we see there all the slopes agree shear goes to 0 so the slope of the moment curve goes to 0 alright any questions on those Friday we'll start talking about what all this has to do with how we actually shape the beam the side of the beam cross-section itself for the most part shear just needs enough area and needs an appropriate resistance to shear but the moment's going to take a little bit more of our little bit more work and bring in a lot more of the tools that we developed in statics last fall for doing some of these things alright any wrap up questions Philly okay Chris did that make sense now even though yours is a lot better answer yeah won't you come up with that answer but that's great fun I think that's the first time I've ever driven to you that kind of frustration Chris I'm very proud of myself I may have to put that on my resume teaching applications alright that's a wrap