 In order to relate the normal stress distribution, the linear normal stress distribution we derived in the previous video to the internal bending moment, we need to actually look at that internal loading. So we will take a section and look at that cross-section and observe that the internal forces acting on that cross-section will result in only an internal moment Mx. If we now kind of rotate that cross-section we can then plot our linear normal stress distribution sigma z that will be tension on the bottom, compressive on the top, and the resultant of that stress distribution is precisely that moment Mx. So we're going to look at the equilibrium and relate these things. Now the first thing to notice is that there is no normal force acting on that, however we have normal stresses. So although you might say, oh there's no normal force, we actually have to do force equilibrium. We have to ensure that the resultant normal force due to that stress distribution is zero. So we'll do that by looking at the internal force on this small element, this hatched element that we show here, that has a thickness of dy, an area of da, and is located a distance from the neutral axis y. On this small infinitesimal element we have a component of the normal force dfz and that's equal to our normal stress sigma z times da. Now there's no resultant force on the cross-section so if we integrate, if we sum all of those forces which is integrating sigma z over da over the entire domain of the cross-section it has to be equal to zero. If we substitute in equation 3 from the previous video we derived a relationship for sigma z. We had that sigma z is e over r times y. What we see is that at a specific cross-section our Young's modulus will be constant, our radius will be constant, so we can remove that from the integral and we get fx is equal to e over r times the integral of y da. Now the integral of y da should be familiar to you guys from your statics course that is precisely the first moment of area about the neutral axis because y is our distance from the neutral axis. Now if we look at this equation it has to be equal to zero so that actually means that our first moment of area has to be equal to zero because if our stiffness is zero there's no beam it has no resistance. If the radius of curvature is zero there is no bending so those are trivial results. So the only result that makes sense is that our integral of y da has to be equal to zero. Now what does it mean when our first moment of area is zero? That precisely means that our neutral axis that we are using for our datum has to pass through the centroid of the cross-section. So the neutral axis passes through the centroid. That is convenient it helps us locate very quickly that datum that we've been using in all of our analysis because we always specify y as the distance from the neutral axis. So that completes the force equilibrium it allows us to establish that the neutral axis must pass through the centroid. Next we want to look at the moment equilibrium. In this case the moment due to area da will be the infinitesimal force df times the moment arm y and we can replace dfz by sigma z times da as we did in the previous slide. Now if we sum all of these differential moments or infinitesimal moments we have to integrate sigma z times y times da and that has to be precisely equal to our internal moment m or mx as shown on the right. If we again substitute in the result for sigma z we get e times r times y sorry e over r times y so then we end up getting e over r pulled out of the integral because they're constant at a given section and we get the integral of y squared da. This should also be familiar from your statics course because it is the second moment of area about the neutral axis. Okay and we denote this by the capital letter i. If we substitute in the the notation i for the second moment of area we can get the equation that m over i is equal to e over r. So now we have these two equations from equilibrium and the result of the strain distribution from the previous video. We can now put all of these things together and develop an equation for the normal stress. So we found out in the previous video that sigma z is e times y over r. We found out today that m over i is equal to e over r. If we combine these two we get what we know as the flexure formula or bending formula and that's that the normal stress sigma z is equal to m times y over i. Now I just want to remind you about our discussion we had earlier about coordinate systems and the fact that this definition is dependent upon the coordinate system you use. Obviously if we define y as positive downwards and we have a moment as shown here we'll get a positive normal stress for a positive y. In some textbooks they define the coordinate system differently and this formula may be having a negative sign in it so you need to be careful about being consistent with your coordinates and the formulas that apply for that coordinate system.