Now I want to start Topic 4 in Mechanics of Materials, Shear Forces and Bending Moments in Beams. These are the topics we'll look at here and in particular, in this segment we'll do some introduction, basically what we're looking for here. And then look to the relationships between the bending moments and loads in beams. Now, beams are obviously very important structural elements, they are subject to various different forces, moments, etc. And we have to be able to design them to resist these loads. So, what we'll be doing here is, first of all, a brief analysis or review of two dimensional beam and frame analysis, which is just statics. Then we'll look a little bit at different types of beams and loadings. And then how to compute the shear forces and bending moments in the beams. So we will analyze this to find the relationships between the shear forces and bending moments and the applied loads and moments. And then learn how to draw diagrams to help analyze beams. And later on in another module we will use these to compute stresses, strains and deflections of beams. So, firstly, simple beams might look something like this. This is a simply supported beam, it has a pin support at the left-hand side, a roller support at the right-hand side and is subject, for example, to concentrated forces here, and distributed load here, and possibly concentrated couples and moments. So we will assume that all of the loads and moments and also bending are in one plane. And the forces can be either concentrated, distributed and generally speaking, we can have concentrated moments. So, we can analyze this situation obviously by statics. This situation is statically determinant but the question that we want to look at here are what about the internal forces in these beams? How are they related to the loads? So firstly, we have any loaded beam produces internal stresses and strains. So to find these, we need to find the internal forces and couples on any cross section at any location within the beam. So let's illustrate this by means of this cantilever beam A B, so it's attached solidly to the wall of B, and we have an upward force here of P applied at the left-hand end, A. So what I'm going to do is divide this beam into two. I'll make a cut through here at mn, and then divide the beam into two separate sections with a free body diagram of each section or segment. So firstly, the left-hand section looks like that, and the section to the right of the cut, the right-hand section, looks like that. And I'm assuming that the shear force and the bending moment on those sections are V and M, respectively. Now the stresses are somehow distributed over that cut, they're not necessarily uniform, but I will denote the resultant shear force by B and the resultant bending moment by M. And in this case we can see that P is transverse, it's not in the actual direction, there's no actual force here, therefore, there is no net actual force in the rod. And I say no net actual force, locally there can be forces but the net actual force must be equal to zero for this case. And also, of course, from Newton's law we know that M and V, the bending moment of the shear force must be equal and opposite on both of those cuts. Now if I apply static equilibrium just to this segment here, summing the forces in the vertical direction, we see that the shear force V must be equal to a constant along the beam and equal to the applied force P. And if I take moments, if I take moments about this end here, I see that the moment is equal to the moment of the supplied force which is P times its moment arm X. So the moment varies linearly along the beam is equal to P times X. Now, to analyze the shear force and bending moments, we need to adopt a signed convention and the sign convention for shear forces is as shown here. So, the shear force on the right-hand face, here, is positive if it's downwards and the shear force on the left-hand face is positive if it's upwards. So, therefore, you can see that I've assumed that the shear force is, in this case, above positive. The sign convention for bending moments is like this. So a positive bending moment on the left causes the top of the beam to shorten. In other words, it's in compression. And the bottom of the beam, to lengthen, in other words, its intention. So again, we see that I've assumed here that the directions of these moments are both positive according to that sign convention. And here is the relevant section from the reference handbook. So, let's illustrate that by means of an example. We have a simply supported beam with a pin support at the left and a roller support at the right and two down with concentrated loads of P and 2P, where P=4.1kN. The dimensions are, as shown here and the first question is, the bending moment just to the left of load 2P. In other words, right here just to the left of that applied load is most nearly which of these alternatives? So first of all I'm going to take moments about A here. To find the reaction of B, I need the reaction of B here to solve this problem, and I'll assume that my moments here are positive in the clockwise direction. It doesn't matter what you assume here as long as you're consistent here. I'll just assume that clockwise moments are positive. Si, therefore, solving, you should get this solution. And therefore, the vertical reaction B here is equal to 6.72 kilonewtons is the reaction of B. Next, I'm going to take moments about my cut here, remembering my sign convention. So across that cut, and I'm going to take moments about the right-hand side. So my cut here, my direction of my positive moment is that direction on the right-hand segment. So we have M minus because this moment is in the counterclockwise direction which I'm assuming to be negative. Minus RB times C is equal to zero so M is equal to RB times the distance C which is 6.72 times 1.5, or 10.1 kilonewton meters, and the answer is d. So, in this case we found that the bending moment was indeed positive, which we can intuitively see here, because these forces are going to cause this beam to bend or deform in that direction, a positive direction. Next, we want to calculate the shear force. So the shear force, just to the left of the load 2P at this point, is most nearly which of these? So here's my sign convention. So my positive force here on that right-hand segment is in the upward direction here, V. So applying equilibrium, in other words, sum of the forces in the vertical direction to that, we have our upward reaction here of RB. For vertical equilibrium, we have V plus RB in the upward direction minus because it's downwards 2P is equal to zero. So rearranging the shear force of E is equal to 2P minus RB. P is 4.1 and we've calculated our V. So therefore the shear force is equal to 1.48 kilonewtons and the answer is A. And this turns out to be a positive number. Therefore, our assumption of the direction of the shear force here was correct. It's in the upwards direction. Another example, we have a cantilever beam which is loaded as shown. We have a concentrated load of 4.5 kilonewtons here. And a distributed load of 1.8 kilonewtons per meter here. And the first question is the bending moment 0.5 meters from the support is approximately which of these alternatives? So 0.5 meters from the left is right here, 0.5 meters from the point A. So, first it's convenient to replace this distributed load here by its equivalent resultant concentrated load. So the resultant load here, which I'll call FR is equal to 1.8. The concentrated force per unit length, multiplied by the length, which is three meters, is 5.4 kilonewtons, and by symmetry, it occurs halfway along here, in other words, 1.5 meters from the point B. So now, I'm going to make my cut through here at the point, 0.5 meters from the left-hand side, and remembering my sign notation again, so a positive bending moment is in this direction on that left-handed cut. So taking moments about that point about the cut and again assuming that clockwise moments are positive we have M plus the moment of the 4.5 kilonewton force, 4.5 times its distance which is 0.5 meters. Plus the moment of this resultant force which is 5.4 times its moment on which is 3 meters from the cut is equal to zero. So computing this we find that M is equal to negative 18.5 kilonewton meters. So the answer is B. So, in this case, the direction of the moment here is opposite to what we expected. Again, we can intuitively see this, because these forces are going to bend the beam in this direction. In other words, the top of the beam in this case is going to be in tension and the bottom is in compression, opposite to our sign convention. Next we want to compute the shear force at that same point. So the shear force here, again, I make my cut through here. And remembering my sign convention that shear force on the left-hand face is in the upward direction, so V is in that direction and summing the forces on the beam to the right of that cut, in the vertical direction equals to 0, I get V minus, because this is pushing downwards, 4.5 minus the resultant force, which is 5.4, which gives me a shear force V of 9.9 kilonewtons, and the answer is A. So in this case my assumption about the direction of the shear force was correct. It is in the upwards direction. Another example, we have a beam with a distributed load on the left-hand side. It's simply supported with a pin-support and roller-support again. But now, we have a concentrated moment of 4.5 kilonewton meters applied at the end C. So the question is the bending moment at the midpoint of AB is approximately and the midpoint of AB is right here, which of these alternatives. So, here again, it's convenient to replace this distributed force by its equivalent single resultant force, and the resultant force is 15 kilonewtons per meter multiplied by the length over which distributed is 24 kilonewtons, and by symmetry it acts halfway across here, in other words, 0.8 meters from the support point A. Now I make my cut and I'll take moments, well not yet, but first I'll take moments about B to find the reaction at A, and that is fairly straightforward. So the reaction of A is 19.4 kilonewtons upwards. Here's the reaction at A. Next I take moments about the cut and set them equal to zero. And remembering my usual diagram here the moment about here on the right-hand face is in this direction from our sign convention. So the free body diagram of the left hand, the section of the beam to the left of that cut, I have minus M, because I'm assuming that clockwise moments are positive here, but my positive moment is in the counterclockwise direction, so minus M, plus the moment of the reaction force here, R A times 1.6 is positive, because that's rotating the clockwise direction, minus the moment of the resultant force here. Because that's rotating in a counterclockwise direction, Fr times its moment arm, which is 0.8. So rearranging, I get this expression for M, and finally, the answer is 11.8 kilonewton meters, which is a positive number. Therefore, it is in the direction as assumed here. And the answer is A. Next we want to compute the shear force at the same place. So here the shear force just to the left of that cut here, right here, is approximately which of these alternatives. So again, here's my diagram. And if I'm looking at the left-hand portion of this, my shear force here is positive in the downward direction, according to our signed convention. So, I sum up the forces on the left-hand free-body diagram, in the vertical direction. So, now I have minus V, if I'm assuming that upwards is positive, minus V because V, I've assumed is in the downwards direction, plus the reaction at RA plus FR is equal to zero. So rearranging, I get V is equal to minus RA plus FR or equal to that and the answer is 4.6 kilonewtons and the answer is A, which again, is a positive number. Therefore, my assumption of the direction of that shear force is correct. This concludes our preliminary discussion of shear forces and bending moments.