Hi, welcome to module 16 of applications in engineering mechanics. Today we're going to continue on and do a bending moment diagram, for a multiforce member. 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 are going to do the, the moment equilibrium equation on this free body diagram to come up with a moment relationship. So, I am going to call this right hand side point A. I am going to sum moments about point A. I will choose my sign convention for assembling the equation as counter clock, clockwise positive and I end up with this is clockwise so that's minus M, minus 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 two. And then I have plus M dM. This shear 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, it's orders of magnitude smaller. And you're making it', 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 change, 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. And so in words again, the change in 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 and we have no we, we have a change in the moment, I am 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 the changes from A to B. And the area under that is minus 20,000. So we're going to start of going from 0 to minus 20,000. 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, 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 20,0000. 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 sign of the, 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 a parabola is 2 3rds base times height, plus I add the area of this rectangle underneath, so the total area under that curve is 17,000 to 27,000, so in going from point C to point D, we're going to go up to minus 3,000 and that's a change of 27,000. the value of the slope I mean excuse me, the value of the shear equals the slope of the moment diagram. You can see that the slope is very high here, a value of 11,000 drops down to 5,000. So the a, the slope becomes a little bit less as we go from point C to D. So we start off with a pretty start, steep slope, and then it starts to go until we get to point D. And since we've integrated a parabola now, we know that this shape has to be a cubic, so I'll, I'll, I'll put a three there with a circle, since that's a cubic shape. and, that's far enough for now, what I'm going to do is, is take a break here, and we'll come back Next module and we'll finish up the rest of this bending moment diagram, I'll see you then.