Welcome to back to module nine of mechanics of materials part three. Today's learning outcome is to derive a very important relationship called the moment curvature relationship. And so last time we summed forces in the x direction and came up with the first moment of area being set equal to zero to find the neutral access. Now what I want to do is I want to sum moments about z and the z axis is coming in and out over here. Or left and right on this diagram on the right. And so we've got some of the moments about Z equal 0. And I'll call counterclockwise positive. And let's do that for this little differential piece of element here, dA. And so, it's going to have, instead of the total moment, it's just going to be dM, for that little dA. So I've got DM and it's going to be positive because it's causing a counterclockwise rotation above the axis. And then I've got my force due to stress which is the sigma sub guix times dA into the board, counterclockwise, times its moment arm, which is y, so that's going to be plus sigma sub x times dA, which is the force in the x direction about the z-axis, times its moment arm, which is y, equals 0. And so I have the relationship for this little differential area, dM equals minus sigma sub x y dA. Okay. There we see it again. Now let's sum up over the entire cross section to come up with the total M and so the total M is going to be equal to minus the integral over the entire area of sigma x y dA. But you recall, since we're assuming that we're working in the linear elastic region, that, from the last module, we came up with this relationship, that, for a linear elastic material, stress was proportional to the curvature and varied linearly with the distance, y, from the neutral axis. So I'm going to substitute that in for sigma sub x. So sigma sub x becomes minus e, the young's modulus times kappa times y. Therefore I've got m equals the integral over A. The negative negative becomes positive. E Young's modulus, times kappa, times, now I have y, and y, so that's y squared, dA. Okay, so let's pick up from there. Here I have what I just came up with. And so, let's take the E, the Young's modulus, which is constant, and kappa, which is constant outside the integral. And I will have M equals E kappa times the integral over the area of y squared dA. And that is defined as i which is the area moment of inertia or the second moment of area. Remember the first mode of area was the integral over the area y dA. Second moment area is the integral over the area of y squared dA. And so this again is defined given the symbol I and it's defined as the area moment of inertia. And we'll figure out how to calculate that later on. And it's also referred to as the second moment of area And so I can write this as a little shorter hand now that M, moment, is equal to kappa, or E, actually, Young's modulus times Kappa, the curvature times I. So we're getting a nice relationship here. So there it is shown again. And so the moment curvature relationship then is, that kappa Is equal to 1 over rho or if I solve here, I've got M over E I. And that's my moment curvature relationship. There it is again, M over E I. And we notice now that curvature, kappa, is proportional to the moment. So the greater the moment, the more curvature you get. That makes physical sense, should make physical sense. We also call what's in the denominator here E times I, Young's modulus times the area moment of inertia as the flexural rigidity. And that's the resistance of the beam to bending for a given curvature. So for a given curvature if I have higher flexural rigidity it's going to take more moment to get that curvature so it make sense it's a resistance again the beam bending for a given curvature. And so now we've got this important relationship of moment curvature and we'll be able to continue on. [MUSIC]