[MUSIC] Hi, this Module 3 of Mechanics of Materials Part III. Today's Learning Outcomes are to determine the Internal Shear Forces and Bending Moments in multiforce members and to sketch what's called a Shear Force Diagram for a multiforce member. Now if you recall and did my earlier course applications in engineering mechanics, we actually did this subject in that course. So, what I'm going to do now, is to have you review, or if you're looking at it for the first time, go through this material from applications in engineering mechanics. So, first of all, I want to, look at this beam. This is a multiforce member, a beam. When I have external forces and moments applied to this beam, it's going to generate internal forces and moments along the length of the beam and those internal forces and moments within the beam are going to be different as you go along at different points along the beam. And it's going to be very important when you later, in later course work, design this member or this multiforce member or beam, to know what the shear is at particular points and what the moment is at particular points. So we come up with a graphical tool to show those internal shear force and bending moments at particular places along the beam and that's the graphical tool we're going to talk about and learn in the next few modules. The Learning Outcomes for today's module are to Determine these Internal Shear Forces and Bending Moments in an internal force members and to Sketch a Shear Force Diagram for a multiforce member. Later on we'll do the Bending Moment Diagram as well. So, we'll begin by looking at internal forces and bending moments in a multiforce member. This can be under a number different conditions. Here's a cantilever beam. So I've got a model of this situation over here. The beam or the multiforce member that I'm interested in, in this case is this yellow cantilever beam. It's fixed on the left hand side so that's the fifth, excuse me on the right hand side, that's the fixed support and on this left hand side, we've got a force up which is equal to p. Now that those external forces and reactions at the base, here, are going to cause internal forces and moments inside the beam itself. And their going to vary along the length of the beam, and as I said before, that's going to be very important when we go to design this beam for certain loading conditions. Let's go back and make a cut. Let's go back and make a cut somewhere along that beam. We'll just look at it generically, and when I do that, I'm going to go ahead and look at the portion on the left. And that portion has to be in static equilibrium just like the overall structure. To be in static equilibrium, we're going to have to have a shear force, which I'm going to label as v, that's equal and opposite the applied force p, and this applied force p is going to tend to rotate the body clockwise and so we'll need a bending moment reaction, as well, that's going to oppose that and keep it in static equilibrium. So that's a good free body diagram of the cut section. If I looked at the other part of the section, the base and the right hand side, all we'll have is equal and opposite shear force and moments on the cut, as shown. We're going to have to come up with a sign convention for talking about our shear force and bending moments. The sign convention will be that, if the shear force is clockwise on the material that we're looking at, then that's going to be a positive shear force. If it's counterclockwise in the material, it will be negative. As far as the bending moment is concerned, for the beam, if the moments are going to cause a shape that looks like this, we're going to call that a positive bending moment. So I call that a smiley face. And this tendency to cause this type of a moment will be a frowny face or a negative moment. So those are the sign conventions we're going to use. Let's take one more example here. This is a generic beam situation. Here, I've got a pin on the left-hand side and I've got a roller here on the right-hand side. And I can have some sort of a generic loading on my beam which I'm going to call q. That's a very generic beam type situation. We're going to look at how to draw the shear and moment diagrams for this type of a structure. So what I do is, I'm going to go ahead and I've got my pin here on the left, my roller on the right, and I've got this generic arbitrary load or force per unit length, q. And I'm going to look at an infinitesimally small element or a differential beam element that's of length or width, if you will, dx, and I can draw a free body diagram of that little differential beam element. It's got the shear force on the left hand side, and I've drawn it positive, and the shear force on the right-hand side, and you'll notice that the shear force is slightly different because the load over that differential length of beam is going to cause a little bit more shear, or a different amount of shear I should say, on the right-hand side. And the same for the moment. I've got a moment on the left-hand side and I've got a moment on the right hand side. It's going to be slightly different on both sides of this free body diagram because of the load that's being applied. So we can go now, and look at the shear forces first, in later modules we'll look at the moments, but for the shear forces, to balance them, we have to look at the equilibrium equation of a sum of the forces in the y direction, set it equal to 0. And so we have V up minus this distributed force, q, q is over a length, dx, so the total force is q times dx. It's down, so that's negative. And I've got minus V, which is down, and minus dV, which is down. So I've got V- q dx- V+dv= 0. I can rearrange that, first of all, I can cancel the V In both sides and I can write -q= x and in words, what that says is, minus the value of the force or the load is equal to how the shear is going to change as we move along in the x direction. So I've written that in words here. The negative value of the load at any point equals the rate of change of the shear in the x direction, or the slope, if you will, of the shear diagram. Here again, I've shown my Beam element. I have shown the free body diagram. I've got my relationship between the negative value of the load and the change of the shear in the x direction. I can rearrange that equation and integrate it from two different points on the beam. Now I can integrate it from a .1 to a .2, anywhere along the beam and when I do that, I find in going from .1 to .2, anywhere along the beam the changing shear over that distance is equal to negative the integral of the load dx. So that says it's the area under the load curve is equal to the change in shear. So again I've written that in words. The change in shear between two points equals negative the area under the load curve, which is represented by this integral. We'll use those two relationships as we draw shear force diagrams in future modules. And that's it for today's module. [MUSIC]
