The next topic I want to look at is stresses in beams. So, in this segment we'll be looking at a number of topics here, and particularly in this first section we'll look at strains in pure and also nonuniform bending. So, ultimately, we want to find stresses and let's suppose that we have a simple cantilever beam with an applied concentrated load P, as shown, then the beam bends into a new shape as shown. So, previously we've looked at how loads create shear forces and bending moments in beams, but now we want to look at what stresses and strains these shear forces and bending moments produce. In other words, how do loads cause beams to bend or flex? So, initially our beam is straight as shown in the top diagram here. But this turns into a curve AB, which is called the deflection curve. And we will use the same coordinate system that we've used previously and we'll assume that the beams are symmetrical about the XY plane, and the loads are also in the XY plane. In other words, two dimensional problems. So, the bending, then is in the XY plane, which we call the plane of bending. The deflection of the beam or the displacement we denote by convention by V in the Y direction. And we will be looking at cases of pure and non-uniform bending. Pure bending refers to flexure or bending under a constant bending moment. In other words, this only occurs where the shear force is 0 because from my previous discussion we saw that dM/dx = V. So therefore, if there is no shear force, then the bending moment is constant. On the other hand, non uniform bending refers to flexure in the presence of shear forces. In other words, where the bending moment is not constant. So, let's illustrate that first of all. I have a simple beam here which is loaded by two couples M0 at the ends. So the shear force of V in this beam is everywhere 0 and the bending moment is equal to a constant is equal to M1. And the bending moment diagram therefore looks like that. So, that is a constant bending moment, so that is what we call pure bending. On the other hand, let's suppose I have a situation like this where I have two concentrated symmetrical loads of magnitude P. And in this case, the shear force diagram looks like that. We have zero shear in the center of the beam. And shear of magnitude plus, minus P on each ends. And the bending moment diagram looks like this. So, in this region, or for this beam, we have a mixture of purely, partly pure and partly non-uniform. So, in the central region here, where the bending moment is constant, we have pure bending and, at the ends here where the bending moment is non-uniform, we have non-uniform bending. So it's a mixture of the two. Next, we're going to look at the stresses and strains that result from pure bending. So firstly, beam curvature. The stresses and strains are related to the local curvature of the beam. So, if I draw that beam much exaggerated, the curvature of the beam might look something like this, and what I'm going to do is consider two points, m1 and m2 on the beam and consider their deflections. So I will assume that S is the distance measured along the beam, or along the curve from m1 to m2, and the normals of those two points, okay? So the normals to the tensions of these two points intersect at this point, which I'll call O prime. So O prime is the center of curvature of the arc from m1 to m2. And of course, this distance from m1 to O prime will generally be a very big distance, because the beam is, hopefully at least, fairly flat and not very curved. The distance m1 O Prime is the radius of curvature which we denote by Rho. And I'll also define something which I'll call the curvature, which we denote by kappa, which is the reciprocal of the radius of curvature 1 over Rho. So we can get the relationships between these variables by simple geometry, by assuming that this is just approximately a triangle here, O prime, M1, M2. We find that kappa is equal to 1 over O, is equal to d(theta) by ds. Where d(theta) is the angle between these two segments. And ds is the distance along there. However, because the radius of curvature is very small, the curvature is fairly flat, ds is approximately equal to dx, the horizontal distance. So that is approximately equal to d(theta) by dx. Now, to find the longitudinal strains, in other words, the relationship to the curvature, we're going to apply what's called classical beam theory, which starts like this. So I'll assume that I have a beam which has some arbitrary cross-section, like this, and frontal view as shown here. And I'll assume that after bending, it's deflected something like that, where my assumption of the direction of the bending moment, follows my positive sign convention as shown. In other words, the top of the beam is compressing, and the bottom of the beam is elongating. So, the fundamental assumptions that we make here are that any cross section perpendicular to the axis of the beam remains planar and flat after defamation. And secondly, cross sections remain perpendicular to the deformed axis of the beam, so these lines here are always crossing at right angles. So a section, or a line which is initially straight, becomes bent and becomes a symmetrical cross-section. Now, the cross-sections here, for example, mn, this cross section, and pq, therefore remain plane, in other words flat, and normal to their local cross sections. And this you can actually prove rigorously, but I won't do that here. So it follows then that all elements, for example, mnpq, this element here, deform identically along the beam. And this is gonna be true for all materials. We haven't invoked any material properties yet. So as this beam deforms, these two lines, mn and pq, rotate with respect to each other. And the angle of rotation between them we will denote by d(theta), and it also follows that mp at the top of the beam shortens, in other words, the top of the beam is in compression, and nq at the bottom of the beam here elongates, it gets longer, therefore the bottom of the beam is in tension, under a positive bending moment. These are normal strains, in other words, linear strains. But in this case, then, the top line is getting shorter. The bottom line is getting longer. So somewhere in between those two there must be a line which doesn't change length. This line right here. And this line where that occurs is called the neutral surface. And the intersection of that neutral surface with the x-y plane is called the neutral axis. Now, to continue with the analysis we have mn and pq intersecting at the center of curvature, O prime. And rho is the radius of curvature to the neutral axis, in other words the axis which doesn't change length under this deformation. So dx, the length along there, along the neutral axis, is therefore equal to a constant. It doesn't change. So Rho d(theta) is equal to dx from simple geometry. Now I'm going to consider a line ef, this line right here. At a distance of y above the neutral axis. So, again by simple geometry from similar triangles, we see that epsilon x is equal to minus kappa y. Where epsilon x is the linear strain of that line EF. So from geometry, epsilon x is equal to minus kappa y, where kappa is the curvature, and y is the height above the neutral axis. And this equation is known as the strain curvature relationship. So, this is purely geometric. And we see then that the longitudinal strain is directly proportional to the curvature. The tighter the curvature, the more strain. And the sine convention if y is positive, in other words, measured upwards from the neutral axis, then the strain is negative, it's compressing. On the other hand, if y is negative, in other words, measured downwards from the neutral axis, the strain is positive. And this follows the same as our previous strain conventions for normal strains. So a positive curvature looks like that. And a couple of notes that everything we've done so far has been based only on geometry. We haven't invoked any material properties yet. And also, we see that the strains vary linearly with height y, regardless of the stress/strain relationships. In other words, regardless of material properties. To find the stresses, however, we do need this relationship between stress and strains, which we'll do shortly. And, finally we also note that this gives rise to strains in the y and z directions. In other words, in the directions in and out of the diagram here, as a result Poisson's ratio. However, it doesn't give rise to stresses in those directions, as the material or the beam is free to deform laterally in that direction. So therefore in uniaxial stress, we have pure bending. We will continue the remainder of this lesson in the next segment.
