[MUSIC] This is Module 13 of Mechanics of Materials Part IV. And today's learning outcome is to resolve that statically indeterminate beam structure. But instead of using superposition techniques for the deflection, in this module, we'll use singularity functions to write the deflection equation and show you that you get the same answer. And so this was our example. I'm going to go quickly through these first few slides because they're identical to what we did in the last module. We wrote the free body diagram, or we drew the free body diagram, we solved for the reactions. We came up with two equations for the static equilibrium, but they had three unknowns. And so we needed an additional equation, which was the deformation equation. And so last time we used superposition techniques, but this time I want to go ahead and write a moment equation using singularity functions that I'll then integrate to find the deflection for this beam or the deformation equation. And so I'd like you to go ahead and write the moment equation using singularity functions, and then we'll integrate it twice to find that deformation or deflection equation. And if you write it on your own, here are the standard forms for the singularity functions for different types of loads. And so I've got our moment equation is a function of x is the moment reaction at the left hand side, it's negative. It's being applied at x equals 0, so we have x- 0 to the 0th power over 0 factorial. Then we have a A sub y also being applied at x of 0, it's a point load, so it's raised to the first power times 1 factorial. Then we have 73 kilonewtons per meter down, so it's negative for our distributed load. It also starts at x equals 0. So it's x- 0 now raised to the 2nd power over 2 factorial. Then we have our 45 kilonewton point load applied at x equals 1.8, point load's raised to the first power over 1 factorial. And then finally we have + By for the reaction at the right hand side, times x- 2.4 since it's being applied at 2.4, raised to the 1st power since it's a point load, over 1 factorial. And so that's your moment equation using singularity functions. Remember again, we made the assumption, this uses the assumption that the beam stays in the linear elastic range, and we should check that. So here's that moment equation. What should we do next? Well, we're going to go ahead and first of all, see that B sub y, that term is going to zero out. It's not going to contribute in the singularity function form at the moment because at x- 2.4, what's in the brackets is equal to 0, and so from there, what we do? Well, we can integrate now. Here's the way that we integrate singularity functions, and I'd like you to do that on your own and come on back. And if you integrate, this is the result you should get. We now have to solve for C1. How can we solve for C1? Well, we can use the boundary condition. We know that the slope of our beam has to be 0 at the left hand side. And so if we put in 0 for all these values of x, we find out that C1 has to be equal to 0, because dy dx = 0 at the left hand side. Again, we integrate, here's the next integration. How do we solve for C2? Well, we're going to have to apply another boundary condition. And that boundary condition is that y at point 0, the deflection at point 0 is also equal to 0. And so we get C2 equals 0. And so now, this is the equation that we're left with for the deflection. We want to find the deflection at, or excuse me, we gotta use the other boundary condition for the deflection to come up with another equation that's going to be our deformation equation or compatibility equation, that we can use with our static equilibrium equations. We also know that y is equal to 0, that's the deflection is equal to 0 at an x of 2.4. And so if we put that in, we can see now, if we calculate that out, we now get a third equation, which relates Ay and the moment reaction at point A on the left hand side. And so now we have three equations, two due to static equilibrium, one due to the deformation equation. We can solve three equations simultaneously, and we get the exact same results we got using the superposition technique, which we must. And so, another way of solving statically indeterminate problems. And so you should have a pretty good handle on this at this point. And we'll see you next time. [SOUND]