[MUSIC] Welcome to module 23 of mechanics of materials part three. Moving along in the course, we just finished inelastic bending. And even though we have bending, we've looked at pure bending we're now, we're going to look at the fact that during bending we also have shear stress in beams. And so, we're going to be able to calculate that shear stress. And so today's learning outcome is to derive an expression for the shear stress in beams that are subject to non-uniform bending. And so for our bending theory, we said that the, our pure bending theory, we said that the flexure was under a constant bending moment, and so there was no shear force present. But we can't have non-uniform beam bending so we get beam bending but in addition to the bending due to this non uniform or this force per unit length being applied to our beam we also get shear stresses. And so we get flexural bending moments in the presence of shear force and also the moment changes now along the length of the beam. And so, let's go ahead an take out a small differential element of that beam and draw free-body diagram. In fact, we drew this same free-body diagram when we looked at shear moment diagrams back in my course, Applications in Engineering Mechanics. And I refer you back to that course if you're having difficulty with that. That's modules 14 through 17 in my Applications in Engineering mechanics course and so here's our differential element with the forces acting on it and forces in moments acting on it. The neutral axis will be located and then we're going to restrict this to elastic regent. And so now let's take a cut to find out if there are shear stresses present, and what those shear stresses are. And so we know that the distance down to the extreme fiber is c, and we'll call the distance to wherever we want to make our cut the distance y. And now I'd like you to go ahead and draw a diagram of what's below the cut. And all I want you to do is show forces in the x direction, because we're only worried about shear in that x direction. And so when you do that, you've got stresses on either side. When you multiply them by areas you're going to get forces. And because the moment changes as go along, the stress and the corresponding force on the right hand side is going to be different than the left hand side. If we assume sigma2 is greater than sigma1, we're going to have a sheer stress that's going to go back to the left. And so sometimes it's hard to visualize where this sheer stress is coming from looking at this diagram. Add so, by looking at, let's look at a beam that is composed of three meet layers. And if I go ahead and subject that to bending you can see that there is a tendency to want to have the section slide relative to one another. And this is where that shear stress comes along and so that's a visualization that should help you know why the shear stresses are generating. And so here's our free body diagram. Let's go ahead now and sum forces in the x direction and so do that on your own and come on back and see how you did. So I'm going to call to the right positive. And I've got my stress over the area, and in this case I got sigma two going to the right, d area, and the area is going to be taken from the location y out to the location c. And the same thing for the righthand side it's going to be negative but we'll take it to the righthand side. So we have the integral from y to c of sigma1 times dA and then we also have now going back to the left, so it's on the righthand side. My shear stress times the, width now b times dx, so that's the area over which the sheer stress acts. Okay and so b is the width. We know now, we found before, the magnitude of the normal stress for linear elastic material due to bending. So we found that sigma was equal to My over I, where, again, y was equal to the distance from the neutral axis. So let's substitute that in. You'll notice that for sigma2 it's M plus dM because the moment can change along the length of the beam. Substitute for sigma 1 as well and here's that substitution shown again. And if I multiply it out, this is the result I get. I can cancel these two terms because they're equal. Let's look at this term, the integral from y to z of ydA. You'll recall that when we went to locate the neutral axis we set the integral of the area, ydA over the area equal to 0. But in this case, the integral were not in neutral axis, so the integral from y to z of ydA Is what we call the first moment of outward area, and we give it the symbol Q. And you'll learn how to calculate that as we go through some problems. And so here's a result. And so we end up with dM/I, times what we call the first moment of outward area, equals the shear stress times the width times dx for this infinite decimal section and tau. Now I can solve the tau and this is the result I get which you will recall again back from my force of bending moment module which we did In applications in engineering mechanics that the change of the bending mode, dM dx, between two points equals the area under the shear force curve. So this is the equation we came up with, I can now substitute that in. And so I have this as V. And I can solve now for tau that's what we wanted to do. We wanted to find what this transfer shear force is. And so it's going to be equal to the V, the shear force. Excuse me, we want to find what the shear stress is, the transfer shear stress. It's V, which is the shear force, times Q, which is the first moment of outward area, divided by the area moment inertia and the width. And so that sheer stress for a non-uniform beam bending will use it to actually solve some problems. And we'll see you then. [SOUND]