[MUSIC] Hi, this is module five of Mechanics of Materials part III. In today's learning outcomes, we're going to, once again, determine Internal Shear Force and Bending Moments for multiforce members. But in this module, we're going to start to look at sketching a bending moment diagram for a multiforce member. And that topic was also covered in my earlier course, Applications in Engineering Mechanics. So let's take a look back. >> So here is my generic beam again and my differential element that I've taken out of the beam and drawn my free body diagram. Now we're going to do the moment equilibrium equation on this free-body diagram to come up with the moment relationships. So I'm going to call this right-hand side point A. I'm going to sum moments about point A. I'll choose my sine convention for assembling the equation as counterclockwise positive. And I end up with, this is clockwise, so that's- M,- V times its moment arm, which is dx, plus I have a q times dx force down. So that's going to be positive in accordance with my sign convention. Its moment arm is going to be dx over 2, and then I have plus MdM. This sheer force on the right-hand side actually goes through point A, and so it will not cause a moment. Now, we can neglect this higher order term because dx is infinitesimally small. And when you have a very small value, the square of that value or if you multiply that value together, its orders of magnitude's smaller. It's a very negligible compared to the rest of the terms. And so by doing that, I can then cancel M and M, and I come up with a relationship that the value of the shear at any point is equal to the change of moment in the X direction. So in words, the value of the shear force equals the slope or the rate of change of the bending moment diagram at any point. We can take that relationship then, and again, integrate between two points on the beam. And when you do that, you find that the change in moment between two points of the beam is equal to the area under the shear curve in this case, as we integrate between those same two points. In words again, the change of the bending moment between two points equals the area under the shear force curve. So now let's use those relationships to actually do the bending moment diagram on this example, we started last module, we had the loading situation. We came up with the shear diagram. On the moment, again, we're starting from the left here and moving right, and so we've got zero. We always start off with a zero bending moment, the end. We have a change in the moment, I'm sorry, between points A and C. Because we do have area under the shear force diagram between those two points, and I've calculated the two points. Actually, the first point we have a change is from A to B. And the area under that is minus 20000, so we're going to start off going from 0 to minus 20000. And the slope at any point in there is equal to the value of the shear itself. It's constant, so it's going to be a constant slope, straight down, it'll be a ramp. And so, that's a straight line, so I'll label it with an s. Here at b I do have to stop, because at b, I do have an applied moment to my beam. And that applied moment is going to cause a smiley face on the beam to the right, and so that's positive. And so, we're going to go up a value of 10,000. So if I go up 10,000, that's going to put me at minus 10,000, and that'll be straight up. And then going between point B and C, I have area under the shear curve of minus 20,000, so I'm going to drop down again another 20,000. So I'm going to drop down to minus 30,000. And the value of the shear is constant, so that slope is constant going down, and that's going to be another straight line. Then in going from point C to point D, the change in moment is equal to the area under the shear curve. Now instead of going down as far as being negative for the shear, this shear is on the positive side of the axis, so this is going to be a positive shear. It's 12,000, which is the area under the parabola here. The area under parabola is two-thirds base times height. Plus I add the area of this rectangle underneath. So the total area under that curve is 17,000, excuse me, 27,000. So I'm going from point C to point D, we're going to go up to minus 3000. And that's a change of 27000. The value of the slope. Excuse me, the value of the shear equals the slope of the moment diagram. We can see that the slope is very high here. A value of 11,000 drops down to 5,000, so the slope becomes a little bit less as we go from point C to D. So we start off with a pretty steep slope, and then it starts to go until we get the point D. And since we've integrated the parabola now, we know that this shape has to be a cubic. So I'll put a 3 there with a circle since that's a cubic shape. And that's far enough for now. What I'm going to do is take a break here. And we'll come back. Next module and we'll finish up the rest of this bending moment diagram, so I'll see you then. [SOUND]