Hi. This is Mechanics of Materials Part 4. This is Module 10, and the learning outcome is to solve for the beam deflection using the superposition techniques that we talked about last time. We said that for a beam that was bending in the linear elastic range, we could use superposition to look at the resultant effect of several loads acting on the member at the same time. So we'll go back to a real-world example that I did in my third part of this four part series of mechanics materials courses. This is a typical bridge. We took a slice of that bridge, may be one bridge girder and we looked at what the loading would be, a typical loading that we would see. We modeled it as shown here. We're going to say that the beam is made of steel, and I give you a modulus of elasticity for the steel, and I'm giving you an area moment of inertia of the cross section of the steel. So you'll also recall that for this type of loading, we have a combination now of a distributed load of 250 pounds per foot, which gives us the deflection as shown here, and we're also going to have a deflection due to the point load at the center which is shown at the right. So the total deflection is going to be a combination or a superposition of those two loading conditions. So if we do that, we know that the max deflection for both of these cases occurs at the center. So it's at L over two, and we've added them both together, there in the negative direction because y is defined as positive upward, and the deflection is going to be downward. If I go ahead now and substitute in the values that I give in the worksheet example problem, this is the mathematics that you'd need to go through. When you substitute in, you'd find that the deflection due to the distributed load would be 3.09 inches down in the center, and the deflection due to the 2,000 pound point load in the center would be 0.33 inches down. So the total deflection would be 3.42 inches down at the center. So that's our answer. So you can see if you have beam tables for common loading situations, this is a very quick and efficient way of finding what the deflections would be. See you next time.