[MUSIC] This is module eight of Mechanics of Materials part III. Today's learning outcome is to talk about how we locate the neutral axis or surface for a cross-section of a beam subjected to pure bending. And so here's what we did last time, we talked about curvature and radius of curvature. Our beam was under pure bending we came up with a strain curvature relationships. We found out that strain was proportional to curvature and very linearly with distance y from the neutral axis and this was our sign convention and we said that was independent of material when we're talking about strains, but we're going to start talking about stresses this time. So let's start talking about stresses and lets say that our beam has this sort of a T type cross section. Actually, it's kind of a cross between an I beam and a T cross section. And we want to be able to know where is that neutral axis located at. We talked about last time, for this type of beam bending, we're going to have compression above the neutral axis and we're going to have tension below the neutral axis, and I've shown that here. If you recall back to my first course in mechanics and materials, we actually did a normal stress strain diagram, and it looked like this. And we're going to be operating now in the linear elastic region down there, where we have a modulus of elasticity relating the stress to the strain of E or what we call Young's modules. And we found that we had a law called Hooke's law that related and is valid in the linear elastic region that says, that normal stress is equal to Young's modules. Where the modules V elasticity times the strain, again that's in a linear elastic region. And so here is that shown again and we now have for our beam what we developed last time is our strain curvature relationship so I can use that strain curvature relationship to substitute nt to Hook's law, and so I find out that sigma of x, in the x direction is equal to e times the strain. Well the strain in the x direction is equal to minus y over row. Or minus e times cappa the curvature times y. And so, for linear elastic material now, we're talking about material, stress is also proportional to curvature and it also varies linearly with the distance from the neutral axis. And you'll know also that from most materials in the elastic range That it's reasonable to assume that the tension in the compression stress strain-curve is the same. So therefore Hooke's Law applies the same for both tension and compression above and below in neutral axis. And so here we have our strain-curvature relationship. We have our neutral axis but where is that neutral axis? Well let's Look at a little area, DA. So, we're looking at the cross section, now, into our beam. At the top, we're going to have stress going into the board, right? Because it's in compression below, it would be tension out of the board. And we're going to sum forces in this x direction, in and out of the board. And so there's stress on each one of these little areas, dA, and so we have sigma times dA. That's going to be the force in the x direction, and we're going to have to integrate that over the entire cross section, over the entire area. And so I can substitute in, now, for sigma sub x being minus epsilon kappa times y. I can factor out not epsilon but E. E is Young's modulus, it's constant. I can factor that out. Kappa's going to be constant. We can factor that out. And we end up with the integral over the area of y dA equals zero. So what we're doing is by my I'm saying y is positive up. So I'm taking Y times DA for every little piece of area above and below the neutral axis is summing in the map and setting it equal to zero. And where that occurs, that's going to be my neutral axis. Okay, where the integral of Y DA equal zero. Will be the neutral axis. So we call that the first moment of area with respect to the z-axis. Because the z-axis would be in, or out of the board here. Or, this direction here. We're bending about that z-axis and so the integral of ydA is the first moment of area about that z-axis. and that actually is the definition of, that's how we found the centroid before. If you go back to my first course in mechanics materials and review modules 19 through 20, you are going to see that, that defines the centroid about the z-axis. Also the centroid bisymmetry along the y-axis would be this vertical axis here, by symmetry also has to be in the middle. Go back again to modules 19-22 in my introduction to engineering mechanics recourse if you want to review centroids of areas and volumes. We found out the centroids doesn't necessarily have to lie on the body for different cross-sections, but it will lie on the axis of symmetry. That's how we're going to find the neutral axis. We'll do it with actual numbers later in the course with examples, but the neutral axis coincides with the centroidal axis of the cross section. And this again is limited to flexural loading and elastic action, we're in the linear elastic region. [MUSIC]
