Hi, this is module 3 of Mechanics and Materials part 4. Today's learning outcomes will be to review where we derive the relationship between load, shear, and moment, and then extend that to the slope and deflection of a beam, and then to determine the maximum deflection of a simply supported beam with a concentrated load at the center, and where that max deflection occurs. So here was the differential equation for the elastic curve of a beam that we came up with last time. We said that if we have this equation, and we also have the equation for the moment along the beam M as a function of x, we can find the deflection by integrating twice and using boundary conditions to find the constants of integration. So here is a structure, a real-world structure in a testing facility that's undergoing a load and a deflection is taking place. What I'd like you to do to start, is to model and sketch this situation as a free body diagram. This is what you should've come up with. On one end you can see that it's pinned, on the other end we have a roller support, and that we have a load, a concentrated load down at the center of the beam. So now that we have this model, what I want you to do is to solve for the reactions at the ends of the beam. If you do that, you can see those reaction forces here, and you'll recall from module 14 in my applications in engineering mechanics course that we came up with this relationship which said that the negative value of the load at a point equals the slope or rate of change of the shear diagram. So as a review, I'd like you to go back and do a shear diagram for this loading condition on this beam. This is what your sheer diagram should look like. Then if you also recall back to module 16 of applications in engineering mechanics, we came up with this relationship, which said that the change of bending moment between two points equals the area under the shear force curve. So here is our shear force diagram, and we can now find our bending moment diagram. I like you to do that on your own as review and then come on back. So this is what you should have come up with for the bending moment diagram. This is our equation now, our differential equation for the bending moment. It's equal to EI d squared y dx squared. So what we can do is since the shear is equal to the derivative of the moment, we can take this derivative and we now have an expression for our shear. In this case, I've assumed that the flexural rigidity EI is a constant. Then I can take the derivative again of the shear, to come up with an expression for the load. So we now have some interesting relationships, and we can go in the other direction by integrating to find first the slope. Here we have an expression for d squared y dx squared. If we integrate that, we'll come up with a slope which is dy dx, and that's shown here. So what I want to do now is I want to come up with what we're going to call the slope diagram. The slope diagram is going to be the area under the moment diagram, will be the change in slope. So you can see the area from this point to the center, under the moment curve is PL squared over 16 because its a triangle. The slope of the slope diagram starts off at zero. We also see if we look at our loading condition that the slope at the center has to be zero, because we're down at the bottom of the deflection. So we know that the slope at the center is zero, we know that the slope of the slope diagram is zero at the left-hand side, but then it increases because of the area under the moment curve to PL squared over 16, so we start off with minus PL squared over 16 EI, and we go to a slope of zero. Then we have a positive slope beyond that, going down to a slope of our slope diagram zeroing out at the end with again, the change in the slope being the area under the moment curve, which is PL squared over 16 EI. Then we can do the same thing for the deflection diagram. The area under the slope will be the change in deflection. Here's our equation, we've integrated again. This is a parabola, so the area under the parabola is shown here, is PL squared over 48 EI. As far as deflection is concerned, we know that the deflection starts off at zero on the left-hand side, and zero on the right-hand side. So we're going to start off with our deflection diagram with zero on the left-hand side. Its slope is going to be a negative value, the slope of the deflection curve is a negative value, and it's going to change down to a value of minus PL squared over 48 EI, and then it's going to have a slope of zero in the center, but then it's going to have a positive slope at the end, ending up with a total deflection of zero at the end. So this is what the curve looks like. We've integrated parabolas here, so we have a cubic equation here, and we now have an expression and a diagram of the deflection of this simply supported beam with a concentrated load at the center. So as a final slide here, we have the maximum deflection of a simply supported beam with a concentrated load at the center is PL cubed over 48 EI, and we see that it occurs at the center of the beam. So that's where we'll leave off on this module.