[MUSIC] This is Module 25 of Mechanics of Materials III. The learning outcome today is to go ahead and solve for the shear stress in a beam subjected to non-uniform bending. And so this is the theory we came up with. There's our expression for Transfer Shear Stress, our definition of outward or first moment of outward area. Last time we discussed how you would find that first moment of outward area. And so now, let's look at a problem and we'll go back to the problem of the beam bridge supporting a truck. And we took a slice, took a look at one beam. And this was the load that we estimated. And so we're going to start this time with a rectangular cross section. We'll go to a different cross section in future modules. But for this rectangular cross section, we're going to say that the beam is steel, subjected to the loading as shown and I want to find the maximum shear force and where it is located. And then find and solve for the transverse shear stress at the neutral axis and at point C on the cross section which is 6 inches below the top. And so the first thing we need to do is find that maximum shear force. And where it's located and to do that you have to do your shear force diagram. And you can do that on your own, come on back and see how you did. And so we see with our shear force diagram that the maximum shear force occurs either on the left end or the right end and it's got a magnitude of 23,750 lb. And so, here's our expression for transfer shear stress. We just found that the maximum shear force will occur at the ends and this is its value. We know that the width of our cross section is 10 inches. We can calculate I is for a standard rectangle, one-twelfth bh cube and there's the value we get for it. The last thing we need to do is calculate the outer areas. For point a, we're looking at the first moment of outer area at the neutral axis. And so here's the area below the neutral axis, y bar to the center of that neutral axis of the outer area is 12.5/2 times the area itself. Which is 781 inches cube. And so, tau substituting in becomes 142.5 psi. And that's our answer to part a, which is the transfer shear stress at the neutral axis. Lets continue on now with part b to find the transverse shear stress at point c. Q the first moment of outward area at c, lets take our cut at c. Look at the area above the cut. And when I do that, I've got the y bar as the distance from the neutral axis of our entire cross section to the neutral axis of the outward areas, that's 12.5- 6/2. And that's times the outward area, which is (10)(6), and we get a value of 570. And so now I can substitute those values in to find my transverse shear stress at point c is 104 psi. Okay, let's redo that and look at our outward area as being below the cut, because it should, as I said, it will be the same answer. And so looking at the outward area below the cut, I've got the distance y bar from our neutral axis to the neutral axis of the outer areas. 12.5, which is this total distance,- 19/2, which is the neutral axis for the area below the cut, times the outward area below the cut, which is (10)(19). And that again comes to 570. So we're going to get the same result 104 psi and so that's a good example of how to find transverse shear stress. And we'll come back next time and look at a cross section other than a rectangle. [MUSIC]