 Let's start off by defining what we mean by a symmetric beam. So take this generalized symmetric beam here with the following cross-section illustrated on the right. Now for a generalized symmetric beam, we are going to make a few assumptions. The first assumption is the cross-section can be any general shape, however the symmetric aspect comes from the fact that the cross-section of the beam will be symmetric across the plane in which the deformation occurs. So our beam will bend in the y direction, so we need symmetry of that cross-section along the y axis. The second assumption we're going to make is that the beam has the same tensile and compressive stiffness. The Young's modulus is the same in both tension and compression. And the reasoning for this will become apparent in a little bit, but for now that is our two assumptions for a symmetric beam that we will use to derive the stress and strain distribution. What we need to do is think about how will such a beam deform. And we'll do that by looking at the simplest symmetric beam, and that is a beam with a rectangular cross-section as shown here. So on this beam I have a series of lines and then a red and a blue line as well indicated on it. And we're going to subject it to a moment Mx. And what I'd like you to do is think about how that will deform. If we look at a video of the deformation of that beam, what we see is that the bending results in a curvature of the beam. We pause a frame out of it and look at that. We can make some observations about the deformation we see. The first observation we can make is that these straight lines that align with the cross-section, although they rotate during the deformation, they remain straight. So we say cross-sections remain plain. Next observation we can make in this particular snapshot of it is that the red line seems to be compressed. It is on the concave curvature side and it's compressed to form that curvature. Conversely, on the convex curved side of that beam, the blue line has to be elongated. So there must be a state of tension in that blue line. And because we have compression on one side and tension on the other side, there has to be some point, although it may not be exactly there, in this case I know a priori it is, but we know there must be a line or a fiber within the beam in which there is no elongation, no deformation. So what we're going to do is take those observations and try and translate it into a description of the strain and then stress distribution in our beam. So if we take our simple beam and we replace it with a schematic of our blue fiber, our red fiber, and a dashed line for this fiber we will call the neutral axis where no deformation occurs. We're going to start labeling a few things on this. First of all, we'll take the neutral axis and we'll measure our radius of curvature relative to that position. Then we have the length of the arc length swept by that radius being ds. So as the radius sweeps through an angle which we'll actually call d theta, our d theta will sweep out an arc length of ds. Now the convenience of measuring r with respect to the neutral axis is we say that the neutral axis does not deform. So ds in our deformed state is precisely equal to the undeformed length dz. Therefore we can reduce this formula to 1 over r is equal to d theta over dz. So this is our first equation that we will use to describe the deformation. Next we're going to look at what's happening at locations other than the neutral axis because we already know no deformation occurs, no elongation or compression or contraction occurs along the neutral axis. So we're going to look at a position of distance y from the neutral axis. So in this case along fiber ef and look at what kind of deformation occurs there. As illustrated here we can see that ef is on the convex side of the beam. So we have an elongation but we'll derive this in a general way. So we look at fiber ef and the length of fiber ef is going to be the instantaneous radius of curvature at fiber ef times d theta. Now we can actually still use our radius of curvature relative to our neutral axis r and just add our distance y from the neutral axis to that to get ref. Taking the equation from the previous slide we can substitute 1 over r dz for d theta and get the following expression dz plus y over r dz. Now if we recall that a normal strain, an elongating or contracting strain is equal to a change in length over an initial length we can say that it is equal to the length of our fiber ef minus the initial length which happens to be dz and divide it by dz. That would give us our strain. So if we substitute these results in we get that our strain in the axis of the beam in the z direction is equal to y over r. The strain is linearly proportional to the distance from the neutral axis. If we plot that we see that we get this nice linear strain distribution with the distance from the neutral axis. Tension on one side compression on the other. Now if we recall Hooke's law in the z direction we have this generalized version of it but in bending we have no normal stresses acting in the x or y direction and we have no temperature change so we can eliminate all of these and simply calculate that the normal stress in the z direction is Young's modulus times the strain y over r. So we get a similar linear stress distribution. So in summary we looked at the deformation of a beam due to a bending moment and saw that we have elongation and contraction strains. These are a result of normal strains within the beam cross section and the distribution of these strains are in fact linear with the distance from the neutral axis. Using Hooke's law we can then also prove that the stress distribution the normal stress distribution must also thus be linear and that is the stress and strain distribution in a beam due to bending. Thank you.