Hi, this is Module 4 of mechanics materials Part 4. Today's learning outcome is to use the double integration method to determine the equation for the deflection of a beam. Previously, we've seen that we came up with a differential equation for the elastic curve of a beam, and we said that if we now have an equation for the moment along the beam, we can find the deflection by integrating this equation twice and using boundary conditions to find the constants of integration, and that's what we're going to do in this module. This is a worksheet where we're going to assume that E and I or the flexural rigidity of the beam is constant. We're going to want to find the deflection of the beam as a function of x and then determine the maximum deflection and where it occurs. So here's a demonstration of the system. Instead of having a moment at both ends, we're just going to put a moment on the right hand side where we have our roller constraint. Okay. What I'd like you to start off with is to solve for the reactions at the pin and the roller in this situation and then come on back. These are the reactions that you should have gotten at the pin and the roller. Now that we have that, what I'd like you to do is to draw the shear diagram. Here's the shear diagram as a result, and draw the moment diagram. Here's the moment diagram that you should've come up with. Okay. Now that we have the moment diagram, what I'd like you to do, we need an equation for the moment as a function of x, and for this moment diagram, go ahead and write an equation for the moment as a function of x. This is what you should have come up with so that we see here at x equals 0, we have a moment of 0, and then it's a ramp function where at the end where x equals L, the moment is equal to M sub End. Now we have our moment curvature relationship and we have an expression for the moment as a function of x, and so what we're going to be able to do is to find the deflection by integrating it twice and using the boundary conditions to solve for the constants of integration. So here is our expression. Let's go ahead and integrate once. So we have EI dy dx is going to be equal to M sub End over L, and the integral of x is x squared over 2. Plus, we're going to have a constant of integration which I'm going to call C1. Then, we're going to integrate again to find the deflection y. So we have EI y is equal to now M sub End over 2L, and if I integrate x squared, we get x cubed over 3 plus now C1 times x plus C2. So we've integrated twice to find the deflection and we've introduced two constants of integration. Here are those results. What are the boundary conditions that we can use now to find the constants of integration? What you should say is, well, we know the deflection at the left hand is equal to 0, and the deflection at the right hand where x equals L is equal to 0. So here's our expressions. Let's use the first boundary condition y of 0 equals 0. That means that x is 0. This term drops out, x is 0, this term drops out and we're left with C2, and so we find out that C2 is equal to 0 as one of our constants of integration. Now let's use the other boundary condition, EI y at x equals L equals 0. So that equals M sub End over 6L times L cubed plus C1 times L, and then we know that C2 is equal to 0. So we can solve now for C1. C1 equals minus M End over 6, and then we've got L squared but we have to divide by L, so it's just going to be L. So we get minus M End times L over 6 for C1. So here's our expression for the slope now. EI is equal to this, and then this is C1. But we want to find as our worksheet solution the deflection of the beam as a function of x. So that would be integrated, and we find out that this is the equation for the deflection. So that's our answer. We'll see you next time.