Continuing our discussion of Structural Analysis, now I want to look at topic II, Beam Deflections. And here, we will first look at the differential equations, which govern beam deflections. Then see how to solve them. Then do some examples. And then look at another method of solution, the method of superposition. So in this section, we'll look at the governing differential equations. Now, we have beams which are subject to various different loading types, and as a result of these loading types, they bend, or deflect, or flex. And previously in Mechanics of Materials, we showed how to compute the internal stresses, and strains. And now, I want to show how to predict the shape of the bent beam. In other words, the magnitude of the deflection. So, the beam may withstand load, but the deflection might be unacceptable, that is why we are interested in this topic. There are two methods we will look at to do this. We will first look at solving it by integrating the differential equations, and then look at the method of superposition. So, first let's look at the differential equations. So, let's suppose that we have a straight beam, which is subject to lateral forces, such as shown here. Then the beam is deflected into a deflection curve. And we denote the deflection in the y direction with the y as the vertical access as V, lowercase v. And also the deflections in the x, and z directions by convention, we denote by u and w. But we don't need to be concerned with those here. Because we will assume that the x y plane is a plane of symmetry, and all of the loads act in this plane, which is called the plane of bending. To analyze this, we'll consider a detail of the curve, which is similar to what we did in the Mechanics of Materials, section VA, as shown here. And blowing it up, a segment of the curve looks like this. So, I'll consider two points denoted by m 1, and m 2 here on the beam. And I will take s as being the distance, measured along the beam in this direction here. So, the normals at these two points, the normals to the tangents intersect at some point, which we denote by O prime. Where O prime is the center of curvature of the curvature of the beam, at that point. And generally speaking of course, this distance from O prime to m, will be a very big distance. Because the beam is relatively flat not as shown exaggerated here. The distance m 1 O prime is denoted by rho, where rho is the local radius of curvature. And another parameter which is useful is the curvature, which is the reciprocal of the radius of curvature that we denote by kappa. Now, we do a little geometry. And by looking at the geometry of this triangle here, n 1, O 1, m 1, m 2, O prime, we see that kappa, which is equal to 1 over rho, is equal to d theta by d s. Where d theta is the angle subtended by those two radii. And we need to adopt the sign convention, and the sign convention for positive curvature is shown here, which is similar to the directions that we get under a positive bending moment. And just as a reminder, the convention for shear force is shown here. So to continue this, we have the local slope of the deflection curve, which is d v by d x is equal to tangent of theta. Or conversely, theta is the arc tangent of d v by d x, the local slope. So for small deflections, we have the distance d s is approximately, equal to the horizontal distance d x. So therefore, kappa which is 1 over rho is equal to d theta by d x. And theta is approximately, equal to tangent theta for small angles, where theta is in radians, is therefore equal to d v by d x. Then we have d theta by d x is therefore, equal to d 2 v by d x squared, or kappa, the curvature is equal to d 2 v by d x squared. Now, we also remember Hooke's law from the section on Mechanics of Materials said, that the curvature is equal to the local bending moment divided by E I, where E is the modulus of elasticity, and I is the moment of inertia. And the product E I is known as the flexural rigidity of the beam. So, combining these things together, we find the differential equation for the shape of the curve, d 2 v by d x squared is equale to M over E I. And this equation is known as the differential equation of the deflection curve. We can extend this by using the local forms of the shear force, and bending moments that we did in the last section. d M by d x is equal to the sheer force, and d V by d x is the negative of the distributed load intensity. To derive these equations for a prismatic beams, in other words, E I, the flexure rigidity is constant along the beam, and this equation we had already. E I d 2 V by d x squared is equal to the bending moment. That's called the bending moment equation, where a reminder of the sign convention for positive bending moments. Next, using the shear force equation here, d M by d x equals V. This equation become E I d 3 v by d x cubed is equal to V, the local shear force. This equation is the shear force equation, with this sign convention for positive shear forces. And finally, the local equation for the variation of shear force is equal to the negative of the load intensity. We get E I d 4 v by d x to the 4th is equal to minus q. This equation is the load equation. Where a positive load intensity is downwards. So, these are the differential equations, which govern the curvature of the beam. And in the next section, we'll see how to solve these, for different situations.
