I want to continue our discussion of stresses in beams. And here we look at the actual stresses and the relationships to moments. Which is the moment curvature equation. And we've used some geometrical properties that we'll need to apply these formally. Firstly, normal stressors in beams. Previously, we've looked at longitudinal strains and we derived this relationship from geometry. That the longitudinal strain, epsilon x is equal to minus kappa y or minus y over rho. Where y is the distance measured from the neutral axis. Now I want to look at stresses that result from these strains. We'll assume that the material is linearly elastic, in other words Hooke's Law applies. That sigma x, the normal stress, is equal to E times epsilon x where E is the modulus of elasticity. Combining these equations, we find that epsilon x is equal to minus E y over rho, or minus E kappa times y. From this then we see that the stress varies linearly with the distance y, measured from the neutral surface or the neutral axis as shown here. And so the stress distribution looks like this. The stress is zero on the neutral axis and increases linearly with distance away from the neutral axis. And a couple of notes about this diagram. That the arrows here are showing the direction of the forces. And the length of the arrows is the magnitudes. In reality of course, the top here is in compression, it's negative and the bottom of the beam is in tension. It's positive, assuming that this beam, is subject to a positive bending moment. Now to use this relationship epsilon x is equal to minus kappa y. We need to know the location of the neutral axis. Where is it? And also, we need to know the relationship between the bending moment M and the curvature. In other words, the relationship between either kappa or rho, and the bending moment and that's what we'll derive now. Now to do this, we note that the stresses have a resultant force in the x direction, the actual direction, and also a moment as shown here. If M is the only applied force, and of course M is not actually a force, than the net force in the horizontal in the x direction is zero. For that element to be in equilibrium. If the beam was a constant cross section, that would mean that these areas are equal to each other. However, if it's not a constant cross section. Then it's something different. But in any case, the net horizontal force must be equal to zero. We get this by summing up the forces on some element here, dA. And the force on that element is the normal stress, sigma x, multiplied by the area, sigma x dA. Therefore, the total force over this whole area here is the summation of the forces on those elements, or the integral of that expression over the area. In other words the integral over A of sigma x dA. But we know what sigma x is? Sigma x is equal to minus E kappa y. Because the force, the stress increases linearly with distance. Substituting that in and under a bending force, bending moment only, the actual force is equal to zero. Therefore, that integral must be equal to zero. E and kappa are constant, so we can take those out. Therefore, it follows that the integral over the area of ydA must be equal to zero. But, the integral of ydA is by definition the first moment and the location then, is by definition the centroid. Therefore, the z axis passes through the centroid of the area. And for the location of the centroids, we can look back on our previous module on statics. This brings us to the very important conclusion that the neutral axis passes through the centroid of the area. But this is only true when the material follows Hooke's Law, and there's no net axial force. Which are the cases that we're looking at here. And remember that generally the y axis for us will be, this vertical y axis here, will be an axis of symmetry. And the y axis also passes through the centroid of the area. Our origin of coordinates here is, the origin of the xy coordinate system is the centroid of the area. Next, we'll relate this to the moment. The net moment is the bending moment which is acting here, and we find the relationship between the bending moment and the stresses. By invoking the relationship that the bending moment must be equal to the sum of the moments on elemental areas. For example, this elemental area A right here. The moment on this element is equal to the force on the element multiplied by its moment on here. Which is, this distance, y. That is therefore, equal to sigma x times ydA and now the total moment must be equal to the summation of the moments on those elemental areas or the integral over the area. M is equal to the integral over M over ADM or substituting we have this relationship right here. And this in turn, because we know our relationship for the sheer stress, is equal to the integral over the area of kappa Ey squared dA. Kappa and E are constants, so we can take those out of the integral, and we get this expression. And this integral here, the integral over the area of y squared dA, is by definition. The moment of inertia of the area about the neutral axis, or around the centroid. Therefore, we see that M is equal to kEI. Where I is the moment of inertia, which brings us to this formula. That Kappa is equal to one over rho, is equal to M over EI. This equation or this formula is known as the Moment Curvature Equation. And the quantity EI in this equation is sometimes called the flexural rigidity of the beam. It's a measure of its resistance or stiffness to bending. Here's our basic relationship. Kappa, which is one over rho, is equal to M over EI. And we already know the relationship between the normal stress and the distance y. And the radius of curvature, which is this expression. Combining those together, we find that the normal stress sigma x is equal to minus My over I. And this formula is known as the Flexure Formula. And the stresses that we calculate here are known as bending stresses or flexural stresses. And we see from this, that the stresses are directly or linearly proportional to the bending moment. And we also see that the bending stress is increased with distance from the new tri-axis. Therefore, the maximum stresses occur at the largest distances from the neutral axis. And if you remember the torsion formula for shear stresses under torsion was this expression. Tau is equal to Tr over Ip. You see that these two equations are very similar or analogous to each other. If we simply replace torque here by moment. And radial distance r with linear distance y, and polar moment of inertia with i, then we have the same formula. They're very similar to each other. They're analogous. Now, to apply these formulas, of course, we need geometrical properties of the cross section of the beam. Particularly its location of its centroid and its moment of inertia. We'll look at those in a minute. And this is the corresponding section from the reference handbook.
