This is Module 14 of Applications in Engineering Mechanics. I want to begin by looking at the overview of the course and seeing how are progressing. We've completed the review of my earlier course, introduction to engineering mechanics. And we went through three of the structural applications. We've done frames and machines, we did plane trusses and we just finished up space trusses. Now we're going to jump down for this next section of the course and talk about internal forces in beams and we're going to look at what's called shear and bent, bending moment diagrams. Then we'll go back, and we'll finish up with looking at cables and structural applications, and finally, the effects of friction on static equilibrium. So first of all, I want to look at this beam. This is a, a, a, a multi-force 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 coursework, design this member or this, multiforce member beam, to know, where the, what the shear is at particular points and what the moment is at particular points. And 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 are going to talk about and learn in the next few modules. So, the learning outcomes for today's module are to determine these internal shear forces and bending moments in inter, internal force members. And to sketch a shear force diagram for a multiforce member. Later on we'll, 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, beam can be under a number of different cond, conditions. Here is a cantilever beam, so I've got a model of this situation over here the, the beam or the multiforce member that I am interested in, in this case is this yellow cantilever beam. It's fixed on the left hand side so that's the fix, excuse me on the right hand side that's the fix 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, at the base here are going to cause internal forces and moments inside the beam itself. And they're going to vary along the length of the beam, and as I said before, that's going to be very important when I go to design this beam. For certain loading conditions. And so let's go back and, 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. So 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 look at the other part of this 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, we're, we're looking at, then that's going to be a positive shear force. If it's counter clockwise on the material, it'll be negative. And as far as the bending moment is concerned, for the beam, if, if the moments that are, if the moments are going to cause a shape that looks like this, we're going to cause, call that a positive bending moment. And so I call that a smiley face. And this tendency to cause this type of a moment will be a frowny face or a, a negative moment. So those are the sign conventions we're going to use. So let's take one more example here. This is a generic beam [COUGH] situation. Here I got a, a pin on the left hand side and I've got a roller here in the right hand side, and I can have some sort of a generic loading on my beam which I'm going to call cue, okay. And so that's a, a very generic beam type situation and look at how to draw the shear and moment dia-, diagram for this type of a structure. So what I do is, I am going to go ahead and I've got my pin here on the left, my roller in 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 with, 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 in the right hand side, adn you'll notice that the shear force is slightly different because the load over that length, over that differential length of beam is going to cause a little bit more shear, or, or 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 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 will look at the moments, but for the shear forces to balance them, we have to look at the sum, the equilibrium equation of the sum of the forces in the y-direction, set it equal to zero. 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 up, minus q dX, minus V plus dV equals 0. I can rearrange that. First of all I can cancel the V. In both sides. And, I can write minus q is equal to dV dx. 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, the rate of change of the shear in the x direction. Or the, the slope, if you will, of the shear diagram. So, here again, I, I, I've shown my my beam element. I've 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. Okay. So I, now I'm, now I can integrate it from a point one to a point two anywhere along the beam. And when I do that I find, in going from point one to point two, anywhere along the beam, the change in shear over that distance is equal to the integral, negative the integral of the load dX. And so that says it's the area of the the area under the load curve is equal to the change in shear. So again, I've written that in, in words. The change in shear, between two points equals negative the area under the load curve, which is represented by this integral. And so, we'll use those two relationships as we draw shear moment, or shear force diagrams in future modules. And that's it for today's module.