Continuing with our statics review, now I want to look at beams and these were topics that we discussed in static section 2B and also extensively in the module on mechanics of materials. We did static analysis of beams under different types of loadings and supports and we looked at the distribution of shear forces and bending moments in the beams. We analyzed the relationship between shear forces and bending moments and the applied loads. And in particular, we saw how to draw a diagrams of shear forces and bending moments to help in our analysis and also to show where the maximum values occurred. Then later, we use them to compute stresses and strains in beams. And now, we will use this to compute deflections and bending of beams. So firstly, let's review some ideas of shear forces and bending moments and we know that a loaded beam produces internal stresses and strains. Therefore, we need to find the internal forces and the couples on any cross-section in the beam. Let's do this for an example of a cantilever beam, which is rigidly connected at the point B. And let's suppose we want to know the shear force and bending moment at some arbitrary cross-section denoted by N, M here at a distance x from the free load P. So to do this, we cut the beam through this location here. We make a cut through M, N and divide it into two free body diagrams. A left-hand and right-hand one. And now the internal forces and moments are the stresses are distributed somehow over the cut. So we will replace those stresses by the resultant shear force, which will denoted by V. And the resulting the bending moment at that cut, which we denote by M. And to remind you of the sign conventions that we use for that, the sign convention for shear forces are shown here and the sign convention for bending moments are shown here. So the shear force on bending moment on each of these faces here are shown in their positive directions. The force P in this case is transverse to the beam. Therefore, there is no net actual force. Although there are internal actual stresses and internal forces, there's no net actual force acting on this beam. M and V are equal and opposite on both cuts from Newton's second law. And to find them, we apply static equilibrium to each of the sections. So by applying static equilibrium to the left-hand cut here, summing the forces in the vertical direction, we see that the shear force V is equal to the applied load P and it's constant along that beam. By taking moments about the cut, we see that the bending moment M is equal to Px. So that increases lineally from zero at the left-hand end towards the support at the rights. Let's do an example on that. I have a cantilever beam, which is rigidly fixed at A, which is subject to a concentrated load of four kilonewtons and a distributed load of 3.5 kilonewtons per meter as shown. Questions are firstly, the shear force at a distance of 0.5 meters from the fixed support is most nearly which of these alternatives? So here, we're interested in a location here, which is a distance of 0.5 meters from A. And remembering our sign conventions then, the direction of the shear force here is in this direction, positive upwards. And the bending moment is in that direction, which we'll look at in a minute. It's convenient to replace this distributed load here by its equivalent resultant and the equivalent resultant force here, if R is the force being at the length times the length. 1.5 x 2 = 3.0 kN and by symmetry it adds midway, it acts midway along that 2 meter segment. So to find the shear force, we sum the forces in the vertical direction on the right hand segment. So here we have V, which is positive upwards minus, because they're both acting downwards, 4 and 3 kilonewtons. So therefore the shear force is positive 7 kilonewtons and the answer is C. The next question, we want to find the bending moment at the same location. It is most nearly, which of these alternatives? So here again, we're interested in the cut here and the bending moment here according to our sign convention is positive in this direction. So, I take moments about the cut and I'll assume that clockwise moments are positive. So we have M plus the moment of the 4 kilonewton force here, which is 4 times its moment on, which is 0.5 meters. And it's positive, because it's acting in a clockwise direction. Plus the moment of the resulting force, which is 3 kilonewtons times it's moment on, which is 2.5 meters is equal to 0 from which we see that M = -9.5 kNm and the closest answer is B. And again, for other similar examples, see section IVa. Now we also looked at the local form of the equilibrium equations or relationships for various loading types. For example, a concentrated load P, a concentrated moment M or a distributed force of intensity q and I will just summarize the results here that they are like this for distributed loads. We have the rate of change of shear force along the beam, dV by dx- q(x) and the rate of change of the bending moment, dM by dx = local sheer force V. For concentrated loads, such as P here. The results were that the change in the shear force, V1 across the concentrated load is equal to -P, where P is assumed to be positive downwards and the change in the bending moment across the concentrated force is 0. In other words, the bending moment is constant. However, the slope of the bending moment, dM by dx is discontinuous across the concentrated load. Finally, we have a concentrated moment of magnitude M0 and the relationships here are that the change in the shear force across the concentrating moment is 0. In other words, V the shear forces concentrated as we pass across there. But we have a sudden change in the bending moment, which is equal to the negative of M0, where M0 is assumed to be positive counterclockwise. So these relationships give rise to rules, which allow us to draw diagrams of the shear force and bending moments, which are especially useful, because they show us the maximum values, which is what were typically most interested in. So the rules we discussed in detail in the mechanics of materials module and I'll just summarize them quickly here. First, the slope of the shear diagram equals the negative of the slope of the sheer force intensity dV by dx = -q(x) from which it follows that the change in the shear force between any two points is the negative of the area under the shear force diagram. The slope of the bending moment equals the local shear force, dM by dx = V from which it follows that the change in bending moment between any two points is equal to the area under the shear diagram. The moment has a local maximum or minimum, where the shear force changes sign from positive to negative or negative to positive. The concentrated forces cause a jump in the sheer force, which is equal to the negative of the shear force, because of our sign convention. In other words, V1 = -P and concentrated moments cause a jump in the moment equal to the moment. Also in the negative direction, because of our sign convention. So these are the general rules and we did many examples of this. Let me just illustrate one here. We have a beam with an overhang, which is simply supported with a concentrated distributed on the load on the left-hand side and a concentrated moment. What do the shear force and bending moments look like for this case? So firstly for the shear force, we have three equations, which help us guide us here and the shear force diagram looks like this. Starting from the left-hand end here, the shear force is 0, then we have a distributed load here of 1 kilopounds per foot. So we have dV by dx = -q(x). Therefore, dV by dx is negative. It's sloping downwards at a rate of one kip per foot. It reaches a minimum value here and then it passes the concentrated load, which is the reaction force RB. This moves it suddenly up to this value, then moving along here, there is no distributed load. Therefore, dV by dx is 0. In other words, the shear force is constant. The concentrated moment doesn't effect that. And then finally, we get to this point, where we have a downward force RC, which is a concentrated force, which brings us back to zero again. Similarly, the bending moment diagram looks like this along with the equations. So at this point at the left-hand end, the bending moment is zero, then we have dM by dx = V. But V, the shear force is negative. Therefore, dM by dx is negative. V is increasing in magnitude, so the slope here is increasing. In other words, this is a curved line here. So this continues until we get to this point here, where the bending moment reaches a maximum magnitude. And the magnitude here is equal to the area under the shear force diagram. In other words, the area of the triangle here. At this point, we have a concentrated load due to the reaction. This doesn't cause a sudden change in the bending moment, but it does cause a discontinuity in the slope. And here again, we have dM by dx = V, but V is now a positive number. So this is sloping upwards and that is equal to a constant, because the shear force is equal to a constant. Then we have a concentrated moment Mz, which causes the bending moment to decrease abruptly at that point. And then finally, it goes back to zero. So in this way, we can readily construct the shear force diagrams to show the variation of the shear force and bending moment along the beam. And particularly, their maximum magnitudes. So for other examples, see the Mechanics of Materials Section, Segment IVb.
