Continuing our discussion of beam deflections, now we will look at how to solve the differential equations. So here are the differential equations again. And to solve them, in other words to find the deflection curve, we have to solve them. And then find the constants of integration by applying boundary conditions. Some examples of boundary conditions are shown here. We have a simply supported beam with a concentrated load, so the boundary conditions are that the vertical displacements at A and B are 0. In other words, VA is equal to 0, and VB is equal to 0. Another example. A cantilever beam with a fixed support at the left hand end subject to a concentrated load as shown. The boundary conditions at A are that this point is rigidly fixed, in other words it doesn't move. So, V is equal to 0. But, also because its rigidly fixed, it doesn't bend at that point. So, the slope DV by DX is also 0. In other words, the beam remains horizontal at that point. And also the second derivative of V is also equal to 0. d2v by dx squared is also equal to 0. So these are boundary conditions and also we have so called continuity conditions. For example that same beam with a concentrated load. To solve this, we would have to solve the differential equations separately for two segments AC and CB. And, the continuity conditions are the continuity conditions for the left and right hand segments at C and those are that the slope here, is continuous. Therefore, the slope of the beam to the left of that point C, dv by dx over AC is equal to the slope of the right-hand portion dv by dx is equal to CB. In other words, the slope is continuous. So, there's no sudden discontinuity there. Similarly, the deflections at that point are also equal. VAC is equal to VAB. The deflection at the point C just to the left of C is equal to the deflection just to the right there. The beam is continuous. There is no sudden discontinuity at that point. So generally we have to solve our differential equations subject to whatever boundary conditions and continuity conditions are imposed by the particular situation. Let me do an example of that. I will integrate the bending moment equation which is EId2v by dx squared is equal to M and I'll do this for this particular case. I can't leave the beam rigidly fixed at A with a distributed load of intense dq and a free end at B. The length is L. So after it deflects, the shape of the beam is something like this. So the maximum deflection I think you can see by inspection occurs at the point B and this maximum deflection we will denote by delta B. Similarly the maximum slope or rotation of the beam also occurs at the point B and this maximum slope I'll denote by theta B. So that's what we're wanting to solve for in this case. So first via simple statics, I find the variation of the bending moment along the beam by taking, let's say, a hypothetical cut at this point, a distance x from the fixed end. And my free body diagram looks like this, which we can derive from regular statics, and you can readily show that the bending moment is given by this expression here. So, now I substitute this into our bending moment equation here, and I end up with this second-order differential equation. And now our task is to solve that equation to get the shape of the beam. So to do that, I first integrate that equation once, which gives me this, E I dv by dx is minus q L squared x over 2, et cetera. And I get there a constant of integration which I'll denote by c1. We find that constant by applying the boundary condition at the fixed end. And the boundary condition is that the slope, dv by dx there, is equal to 0 at x is equal to 0. In other words, the beam remains horizontal at the point, a. And substituting that back into the differential equation, we find that the constant of integration, C1, is equal to 0. So, I can also write that solution, then, in terms of the local slope, dv by dx is therefore given by this expression. From which we see that that is equal to 0 at x is equal to 0, in other words at the point a which of course it should have been. And it's otherwise negative. So the slope is always negative for this case. The beam is bending downwards which is negative, according to our sign convention. Next I'm going to integrate this equation again which gives me this. EIv is equal to minus ql squared x squared over 4 etc. And again I come up with a new constant of integration here which I'll denote by C2. And this constant we apply find again by applying our boundary condition at the point A. And because it's rigidly fixed, the displacement, their v is equal to 0 at x equals 0. From which it follows that C2 is equal to 0. So finally our expression for the deflection v, the deflection curve is given by this expression here. And that gives us the shape of the bent curve. And again, we note that that is equal to 0, at x equals 0, which it must be, because it's rigidly fixed there, and it's otherwise negative. And again, this follows from our sine convention. This is being bent downwards, so the deflection v is negative. Especially we're interested in the maximum values. And firstly, the maximum deflection, in other words, delta b is equal to the deflection at the end of the beam, in other words, v evaluated at L is equal to x. Substitute into this equation right here, we find that the maximum deflection is qL to the 4th over 8EI. Similarly, the maximum slope is equal to theta, or theta b or dv, by dx evaluated at x equals l, which has the solution theta b, the maximum slope, is ql cubed over 6 ei. And that's how to solve for these problems. Now, generally, of course, these kinds of solutions are too complicated to solve in the actual FE exam, where we have less than three minutes for each problem. So this is the extract from the reference hand book, it gives those equations which I just stated. And how to solve them. And then, in the manual, the reference hand book, they have tables of solutions. And I can't reproduce them here because of copyright issues, but it looks something like this. And you can find the table readily. It has a column of label BEAM, which has the different scenarios, in other words, different loading types and different support types. Then it has column labelled SLOPE which gives either the maximum slope and or the slopes at the end of beams. Then it has DEFLECTION. Which gives just the maximum deflection of the beam, and then they have a column which is labeled elastic curve, which is the deflection curve. In other words, the solution V has a function of distance X along the beam. So these will look something like this. These are some examples, not from the reference handbook, but similar examples, showing the solutions for deflections and slopes of cantilever beams, where the maximum deflection is shown by delta b. And the maximum slope, or rotation of the beam, by theta b. For example, the solution to a distributed load, which is the one we just did, is this. Here is the elastic curve or the deflection curve and the maximum values of deflection and slope, or a concentrated load P where the solutions are as given here. Then, we have some examples for simple beams, an example, here, of a distributed load, here are the solutions for the shape of the curve and the maximum deflections and rotations. So the manual gives many examples of these and you will have to, given any similar problem, you will have to find the solutions for that particular case. And we'll do some examples on that in the next segment.