Hi, this is module eighteen, of An Introduction to Engineering Mechanics. The learning outcomes for today are to give examples of body forces and surface forces, to determine the resultant of a force distributed along a straight line and to find and locate the centroid for a force distributed along a straight line. So these are the things you should be able to achieve by the end of this module. First of all distributed force systems are an application of resultants, and we've looked at several modules on resultants, and we've done a, a number of, of problems. So now we're going to get into an application of that where we're going to have distributed force systems. And so the first thing I want you to do, is to give me some examples of body forces. So I'd like you to take a few minutes, and on a piece of paper, write down some examples of body forces. So, now that you've answered that question, here's a, a couple typical examples of body forces. gravity is a body force, and so, if I have a body over here Gravity acts on each piece of mass in the body, and so, or each little part of it, mass in the body and so it acts throughout the body. electro-magnetism is another example of a body force Let's. Also give examples of surface forces, and so, take a few minutes and write down some surface forces and then we'll come back. Now that you've answered that question. Here are some examples of surface forces. One would be, if I was pushing on this book. Okay? This notebook with my finger. That's a force acting externally to the to the, to the body, and so that's a surface force acting of the surface of the body. if there was a huge gust of wind that came through this, this room and the wind was in this direction that wind force would act on the surface of the body, or the surface of the book. And that would be an example of an external force. And you can think of others as well, I'm sure. Okay, let's talk about how to find the result for a force distributed along a straight line. And so here, I hav e, An x, y coordinate system. I've got a force distributed along the x-axis, and it has a varying magnitude as it goes along. And that magnitude is defined by the function f of x, where f is a force per unit length. And what we're going to look for is what the resultant is? What the magnitude of the resultant is and where it acts? And so what we're going to do is we're going to take this total force and we're going to split it up into little differential elements. And so I've written a little differential element here. I've exaggerated the size of it. It would be infinitely small but it has a distance in the x direction or a width of dx and it has A magnitude of f, and so the total force acting down from that little element would be fdx. The force per unit length times its unit length, which is dx. And so, if I want to find the resultant for the entire load here, I take the, the force of that individual differential element fdx, and then I integrate it over the entire force acting from zero to l. And that's going to give me the area under the loading curve, or the magnitude of the resultant force. If I want to find where that resultant force acts. So I found the magnitude here. If I want to find where it acts, this distance x of r. What I do is, I look at the moment due to the re, the, to the distributed force. And that's got to equal the moment Due to the resultant force. So, for the distributive force, each of these little differential elements are acting down at a distance x, so x times fdx, that's going to be the moment. We integrate that from zero to l for the entire distributive force. That's got to equal to magnitude of the Resultant force, which is the integal from zero to l, fdx, which we looked at up here, times it's moment arm, which is xr. And so xr, the distance to the resultant force Its just the integral from zero to l of xfdx and then we divide by the area under the loading curve in the mom the result not the load which is the integral of 0 to l fdx. And then xr becomes the x cordinate of what we call the centroid of this load or the Geometric Center of the Load. So that's the definition of a centroid. Geometric Center. So let's do an example again. Here, I have, up in the right hand corner of my, my definitions. xr is the distance to the resultant force. We just talked about that. F magnitude of fr is the magnitude of the result in force load. And so, here's my example. I've got a simple ramp load here. It goes from zero to b. The function is just a straight line. So the. The force per unit length is B over L, the slope times x. And so I can use my definition x sub r is equal to the integral from 0 to l of xfdx over the integral from zero to l of fdx and so in this case we've got the integral from zero to L, f is bl over x and we multiply that by x so that's going to be B L x squared. DX divided by the integral from zero to L of F which is B over L times X, DX. And so that's going to equal B over L is a constant, B's constant, L's constant. I can pull it out of the integral, so I've got B. Over L times the integral from zero to L of x squared dx. Divided by B over L times the integral from zero to L of xdx. If I integrate those I get X sub r equals B over L times the interval of x squared is x cubed evaluated from zero to L and then B over L times the integral of xdx is x squared over 2 evaluated from zero to L and so that's simply going to be 2 3rds L. So, that's the answer for my resultant location. So, if I have my resultant here, it's going to be at a distance, this is F of R, it's going to be at a distance 2 3rds L. And we still need to find the magnitude of that resultant. So to find the magnitude I have it written here. Magnitude of Fr is equal to the integral from zero to L of fdx that's equal to the integral from zero to L, F is B over L x and that's going to be dx, that's just going to be equal to the area under the loading curve which is the triangle in this case. So if I do the integral, I get the magnitude of the resultant force is e qual to B over L is a constant. Pull it out. Integrate xdx. That's x squared over 2 again. I evaluate it from zero to L. And so I find out that the magnitude of my resultant force is BL over 2. And that should make sense to you because that's the same as the area, that's the area under a curve for a triangle. One half base times height. The base in this case is L the height is B. And so I have my total resultant force, or a single resultant force for this distributed load. It's acting down. It's magnitude is BL over 2 and the distance to that results is 2 3rds L. So, a nice application for a distributed load surface load acting on a body and it's a nice application of results. I'll see you next time.
