Hi and welcome to module 26 of an Introduction to Engineering Mechanics. Today's learning outcome is to continue to use the TD, 2D equilibrium equations to solve for the force reactions and moment reactions acting on a body to keep it in static equilibrium. This is the culmination of the course. And so we're getting towards the end, and we're getting to solve some really neat problems. So we're going to go back to that crane problem that I talked about in the introduction and in the first lesson. So I've got a picture of a crane here or, could be a cherry picker truck or whatever. And so I've drawn a model of that and so I've got this, beam bar part where I have an external couple being applied to raise and lower that. I've got a hydraulic bar here and I can have a load. I've actually drawn the load on my picture straight down but to make it a little bit more interesting I've drawn the load. Out to the left at an angle of 30 degrees, give us a little bit more complexity to the problem if you will. And so we want to determine the reactions at the pin support at A and the force in the hydraulic arm BD, for the truck crane shown, and to determine whether that hydraulic arm is in compression or tension. And we're going to say that these external forces and moments are so large that the, the weight of the members themselves are negligible, or we can neglect them. And so, that's the problem. We want the conditions for that to stay in static equilibrium so we don't want that, that crane to turnover, we'd like it to be stable and so what do we do first? That's, that's my first question for you and I'd like you to think about that and then once you've done that, come on back and we'll talk about what we'll do first. Okay now that you've answered that question, what you should have learned from this course is that the first thing you always want to do is to do this free body diagram, this graphical tool to apply the equations of equilibrium. And so here's a picture of my model again over on the right. And I'd like you to go ahead and draw that free body diagram. So take your time, do it and then let's come back and see how you did. Okay. Here's the free body diagram. We see that we have our 900 pound external force up here at the left. We've got a pin reaction. Oh, I've, I've drawn a free body diagram of, of this bar, this external bar. That the body of interest I identified. I've got a pin reaction at b. So the pin reactions prevents motion in 2 orthogonal directions. It doesn't prevent rotation. So there' s just a bx and a by. I've got another pin reaction at a, ax and ay. And I've got this external couple and I've also put the dimensions on. And so I've got a good free body diagram. So now we apply the equations of equilibrium. The 2 dimensional equations of equi, equilibrium, I have 3 independent equations. But how many unknowns do I have in this problem and what are the unknowns? And so I'd like you to th, to write these down and think about it and see if we're good to go or if we have a problem. Okay, now that you've thought about that we have three independent equations but we have four unknowns. We have the pin reactions at a and we have the pin reactions at b. Ax, Ay, Bx, and By. So I've got a sad face because I won't be able to solve the problem with just those three equations. And so again I want you to think, well what could we could get next, do next. What could we do to get more equations to be able to solve for these unknowns. So think about that, when you've got an answer come on back. So the thing we want to do is to sketch another body another free body diagram, if I can sketch another free body diagram, I will get more independent equations and so I've taken the body of interest for A B C but I haven't done body BD here, this hydraulic arm. So let's go ahead and use member BD. And so, we're going to sketch another free body diagram. And by doing that, I have a pin reaction on the hydraulic arm at point B, and a pin reaction at point D. One important thing that you want to do is make sure that your pin reactions at the top of the hydraulic arm are equal and opposite to the pin reactions that are acting on bar A, B, C. And that's because of Newton's third, third law. If By is acting up on this bar, the pin reactions acting up on bar ABC then it has to act down on hydraulic arm By, same with Bx, Bx is to the right here, would be to the right to the left here. And, then I've drawn my Dx and Dy which are two new unknowns that I've introduced. So, now I have 6 equations. I have 3 independent equations for this body. I have 3 independent equations for this body. And, you've noticed I've only added two more unknowns. So the total unknowns now are the pin reactions at A, the pin reactions at B and the pin reactions at D. And so I have a happy face because I will be able to solve this problem. Okay let's use one of these free body diagrams to start to solve for the unknowns. I've taken the free body diagram of the hydraulic arm BD. First of all, I'll sum courses in the X direction, set it equal to 0. I'll choose the right positive for assembling my equation. And I get Bx is to the left, so it's negative in accordance with my sign convention. Dx is to the right, so that's positive Dx. Equals 0 and so that results in knowing that Bx is going to equal Dx. Let's next apply sum of the forces in the y direction equal 0. I'll choose up as positive. And I get, By is down so that's going to be minus By. Dy is up so that's plus Dy equal 0 and so that equation yields that By equals Dy. And lets just sum moments about any point we choose. I'll choose the sum moments about point B equals zero. I'll go ahead and call clockwise positive for assembling my equation. And I get, if I sum moments about B. I get By its line of action those to point B so it does not cause a moment. Bx's line of action those to point B. So it does not cause a moment. Dy's line of action those to point B. So it does not cause a moment. The only force that's going to cause a moment about point B on this free body diagram is Dx. And it's going to cause a counter clockwise rotation. That's going to be negative in accordance with my sign conventions like that. Minus Dx times its moment arm and we'll just say that the, I, I guess I, I, the, the distance from my diagram was 20 so that's going to be minus Dx times 20 equals 0. And so I find out then that I've solved for Dx. Dx is going to be equal to 0, and since Dx is equal to 0 we also know, from this equation up here, that Bx equals 0. And so I've got a couple of my force reactions done. And so that's where we'll pick up from next time and we'll continue to solve this problem and figure out the rest of those force reactions to keep this crane in static equilibrium.
