I want to complete our discussion of beam deflections by looking at another method of solution, the method of superposition. The method of superposition can be simply stated that the beam deflection due to several different loads can be found by superposing or adding the deflections due to the individual loads. Now the reason we can do that, is that if we look back at the differential equations for prismatic beams, we had these three equations here. These equations are linear differential equations, and mathematically, what that means is that, if we have two solutions to a linear differential equation, a third solution is simply the sum of those two solutions. For us, it means more simply, that we can add or superpose the individual deflections or rotations, due to several different load types. Let me illustrate that by means of an example. We have here a situation where we have a distributed load of intensity Q and a concentrated load of magnitude P, so those two situations are two different cases in the reference handbook. How do we find the total deformation due to those two different load types? Well, firstly, the beam is going to deflect something like this. The maximum deflection occurs in the middle, and the maximum deflection for both of those different load types both occur in the same place, in other words, in the middle. And the maximum rotations also occur in the same places at the ends A and B. So firstly for the uniform case, in other words, the uniform distributed load Q, for a simply supported beam is case 4 in the reference handbook, from which we find that the maximum deflection, which is in the middle, is given by this expression, 5QL to the 4th, divided by 384EI. For the point load at the center, this case, that is case one in the reference handbook from which we find that the maximum deflection is equal to this expression, PL cubed over 48EI. So in this case because the maximum deflections both occur at the same point we can simply add them together. In other words the deflection delta C in the middle is simply the sum of those two individual deflections, and we can do the same thing for rotations. The maximum rotation occurs at the end here denoted by theta A and theta B, so for the uniform case, case four, we have is equal to ql cubed over 24EI. Similarly, the maximum deflection for the point load case, case one, is given by this expression, and because they both occur in the same place, we can say that the deflections at the ends theta a is equal to theta b, is equal to the maximum deflection, is equal to simply the sum of those two individual rotations. And that's the method of super position. Let me do an example on that. We have a cantilever beam of length L, which is acted on by a point load P and a moment M as shown. In this case, we want the deflection at the point B at the end to be zero and for this to occur, the relation between P and M is given by which of these possible equations. So, in this case the situation is that under the concentrated load P, the beam is going to bend downwards like this, so we're going to have a negative deflection. On the other hand, under the action of the moment, it's going to bend upwards, like this, so we'll have a positive deflection at the end. So obviously, for the net deflection at that point to be equal, the magnitudes of those two deflections must be the same. To solve this then, we see that the maximum deflection, for each case, occurs at the end at b. And first for the point load at the center for cantilever beam, this is case eight in the reference manual, from which we find that v max maximum deflection is minus PL cubed over 48EI, and in this case we have to be careful with the signs to get those right. For the bending moment at the end is Case 10 in the reference handbook from which we find that Vmax is equal to positive ML squared over 2EI, therefore the total deflection at the end ias simply the sum of those two because from the method of super position. In other words, v, the defection is equal to that, the sum of those two. However, the question is do we want this deflection to be equal to zero, so I set that equal to zero, and then solve this equation for P, and the answer is P is equal to 24m/5L and so the answer is A. The second part we want the slope at b to be zero, and in this case the relationship between P and M is which of these alternatives? So, in other words, we want the beam to remain horizontal at this point. In other words the rotation's due to the concentrated load in the moment to cancel each other out. So again, the maximum slope for each case occurs at the point B, so we can just add the rotations or slopes for each situation. And for the first case, the point load we find from the handbook that the maximum rotation at the end is given by that expression, which is negative, in other words, it's sloping downwards. On the other hand, the rotation for the moment at the end, which is case 10, is given by this expression, and the total rotation is the sum of those two is equal to therefore that. But again we're given that we want this rotation to be zero, in other words, the beam to remain horizontal, so I set that equal to zero and solve this equation for P with the result that P is equal to 8M over L, and so the answer is C. And this concludes our discussion of beam deflection.