 Let's look at some examples of static equilibrium as applied to a two-dimensional beam. Here's a relatively simple one. Let's consider some form of beam along board. Maybe it could be a diving board or something along those lines. And it's going to be simply supported. So I'm going to place a roller support here and near the left end. And I'm going to place a pin support here, not all the way on the right end, somewhere over toward the right. And we're going to arbitrarily load this beam with a certain number of load forces. Here is a 15k. And in this case, k will stand for kilonewtons or often called kips. This is one of the abbreviations. A small k will be kilopounds or kips. Sometimes you will see kilonewtons, but usually those are abbreviated kn. And we'll also place a six-kip. Let's go ahead and draw the 15 to be significantly bigger than the six. There's a six-kip load on the very end and a six-kip load located partially out on the end of the sort of overhang there. And in order to solve a problem like this, we will also need the appropriate geometries for each of these pieces. We're going to say that there's a three-foot distance between the support and the first load, a six-foot difference between that load and the next support, and then two feet difference between each of the other loads on the beam. So in order to consider this, we now have all the loading that's here. Let's go ahead and think about the free body diagram that's here. The only thing that the free body diagram is going to do is it's going to separate the two supports from the body itself and replace those supports with some reaction forces. Let's go ahead and draw in the appropriate reaction forces. In order to do so, though, we're going to want to make sure we establish a basis first. And the easiest basis for this particular one will, again, be horizontal and vertical. We'll call the x component in the horizontal direction and a y, or perhaps a z component, but here we'll call it a y in the vertical direction. x is horizontal, y is vertical. And let's replace both of our supports with the appropriate reaction forces that go along with those supports. In the case of the roller support, the roller support is unable to resist any sort of horizontal reaction. So it will simply supply a reaction. And we'll call this point A a reaction in the A direction, r, a, y. And we'll call the other support point point B. And notice, because it is a hinge support, we can actually support two dimensions. And we will define those components, r, b in the x direction and r, b in the y direction. So we've established our free body diagram. We know our loads that are known, our 15 kip, 6 kip, and 6 kip loads, and where they're applied. We know our reactions, or we've established unknown reactions, and we've broken them down into components. Now let's go ahead and use equilibrium to solve for the unknown reactions. Our first equilibrium equation is that the sum of the forces in the x direction has to equal to 0. Well, that's fairly simple. There is only one force acting in the x direction, and that's going to be r, b, x, the reaction at point B in the x direction. And we know that that has to be equal to 0. So now we've already solved for one of our unknown reactions. For our second equilibrium equation, we have the sum of forces in the y direction, which must be equal to 0. Well, let's consider all the forces in the y direction. I'm going to move from left to right. First, I see that there is a force, r, a, y, that I've defined to be in the positive direction, pointing up on the far left. And then I have a force of 15 kips, but it's pointing down. So I will make the value negative in this equation. Then I have a force r, b, y pointing up. I have a force of 6 kips, a load of 6 kips that's pointing down, and another load of 6 kips that is pointing down. And that entire sum must be equal to 0. If I simplify that, I have that the sum of r, a, y plus r, b, y has to be equal to the sum of the other three together, which is a total of 27 kips. And if I place that on the far right side, the sign changes. And so we have an equation that relates the two reactions to the loading forces. So we still have two unknowns here. We've used two of our equations, and we've eliminated one unknown. But we have a third equation that we can use here. In this case, our third equation is the sum of moments. And now we want to choose a judicious place to take our moments about. Usually the best idea there is going to be to pick a place that eliminates one or more of our unknowns so that we get an equation that has fewer unknowns in it. So we're going to choose point a to take the moment about. And let's go ahead and define a positive moment, again, as being clockwise around point a. So if I start at point a, I know my sum of my moments around point a. Let's go ahead. We know that's equal to 0. But let's go ahead and sum each of the moments that we have there. First, we see that there is a, well, r, a, y doesn't contribute because it has a moment arm of 0. If we look at our first moment we see is this 15 kips applied at a moment arm of 3 feet. So 15 kips applied at 3 feet. If we keep moving down, we now see that we have r, b, y applied at a total of 9 feet. Notice I'm adding the two together, 3 feet plus 6 feet. 3 feet plus 6 feet is a total of 9 feet. And that's the distance of r, b, y. Also notice that that applies a moment in the other direction in what we're defined as the negative direction. So I'm going to put a minus there. If I continue, I will then see that I have an additional 6 kips applied at 11 total feet, a distance of 11 feet away, adding the 3, the 6, and the 2. And that's going to be in our positive moment direction. And last but not least, we have another 6 kip load applied at the total distance, the total moment arm of 13 feet. And we, again, know that that entire sum must be equal to 0. So let's go ahead and do the math here. We have 45 kip feet minus 9 feet times r, b, y plus 66 kip feet plus 78 kip. And we know that that sum of all those moments is, again, equal to 0. Well, if I do a little algebra and move r, b, y over to one side of the equation and divide everything by the 9 feet associated with r, b, y, I get that r, b, y, the reaction in the y direction at point b is equal to a total of 189 kip feet divided by 9 feet. Well, notice that the units of feet cancel, which we want because we want our answer in some form of force or kilopounds. And the 189 divided by 9 is going to be 21 in units of kips. So that is our reaction force in the y direction at point b. Now we return to our equation for the sum of forces in the y direction, plug in the known value of b, y. So we have that the sum of the a. So we have the reaction force at point a in the y direction plus the reaction force at b in the y direction was a total of 27 kips. Well, if we replace r, b, y with the 21 kips, we get that r, a, y is equal to 6 kips. And that's going to be pushing up in a positive direction. And so we've solved for the reactions in this beam. Notice there's a substantially larger portion of the load being carried at point b, which might be expected, than there is at point a. Let's look at a second example. This one is significantly more complex. In this case, we have something that might represent the arm of, perhaps, a crane. You'll notice that this crane is carrying a certain weight. And that weight is being supported by a long arm. And that arm is being bolstered both by a cable and by some sort of support down at the base of that arm, that beam arm. So we have lots of information about this. We have the angles that the cable are at. We have the angle that the arm is at. But let's start by breaking this problem down. We'll start by creating our free body diagram of our beam. I'll redraw the beam, cut it off at certain locations. Let's go ahead and slice through the cable and replace the support. We'll also slice through and replace the weight with the value that we know is with the value that we know that the weight is. So go ahead and draw these. We'll start by drawing our known loads. In this case, we know we have a weight of 3 kilonewtons. And then we will replace our cable with a tension that we know has to be applied at this angle of 15 degrees. And then we will replace our supports down at the base with two components. We'll call this the reaction at point A in the x direction and the reaction at point B, I mean right at point A, and the reaction at point A in the y direction, if we again establish our basis as being vertical and horizontal. Notice that rAx and rAy, even though there is an angle here and that angle is important of 30 degrees, that rAx and rAy do not necessarily relate to each other along that 30 degree angle. There may be more push in the vertical or the horizontal direction that doesn't work directly up that beam. And that's because this beam is a three force member. Notice there are three forces on it, so it's not guaranteed that this force acts along the beam. So now that we've established all those, we have our known loads and we have our unknown reactions and we've broken one of those reactions into components. Let's break the other reaction into components, but we know how those components are related to each other because that does indeed have to act upon that particular angle, that 15 degree angle. So let's go ahead and look at that tension and we'll notice that there is a component in the x direction and a component in the y direction. And we'll relate those components to the tension itself, that the component in the x direction is equal to the magnitude of that tension times the cosine of our 15 degree angle and the y component is equal to the tension times the sine of our 15 degree angle. We're also going to need a little bit of geometry here to talk about some of the moment arms. If we're going to use our moment diagram, we're going to need to know what our moment arms are. So we'll notice here, first of all, let's redraw the picture here. Here's point A, here's point B, and here's point C, where the weight is applied. And we'll notice that this distance here along the x-axis, the horizontal distance from point A to point B is going to be equal to what we have our L1, this first length of 3 meters. This horizontal distance is going to be L1 times the cosine of our 30 degree angle. And if I want to consider the length from point A all the way out to point C in that horizontal direction, it's going to be equal to the total length L1 plus L2, again times the cosine of that 30 degree angle. One other piece that's going to be useful, if we consider point B here, is we might need to know the vertical distance there. The vertical distance is going to be equal to L1 times the sine of 30 degrees. And similarly, the vertical distance for point C is going to be L1 plus L2 times the sine of 30 degrees. Those values will be useful in determining moment arms when we do our sum of moments. So now that we've considered the different values in the geometry, let's go ahead and apply our static equilibrium equations. Our first equation, sum of forces in the x direction. What are the forces in the x direction? It appears we have two forces in the x direction. We have rax and tx. So we have rax, which is acting to the right, so it's a positive value. And we have the tension, tx, but we've defined that as pointing to the left, so that'll be a negative value. And we know that that's equal to 0. Let's go ahead and do a little algebra, establishing that rax is equal to t cosine of 15 degrees, replacing the tx with t cosine of 15 degrees, and relating it to the unknown rax. Now we'll do the sum of forces in the y direction. Sum of forces in the y direction. We have a reaction at y that's pointing up. We also have a tension that's pointing up component. Here's the tension component. Here's our ay, and here's our weight that's pointing down. So we can add those three pieces, the reaction force at y plus the tension force in the y direction, and then we'll subtract the weight. And we know that that sum has to be equal to 0. And again, if I do a little algebra, we can simplify things just a little bit. ray plus t sine of 15 degrees is equal to our weight, which is the 3 kilonewtons in this example. Now we still have 1, 2, 3 unknowns, and we only have two equations. We need the third equation, so we're going to go ahead and use the sum of moments. Now we need to make a choice about what point to sum the moments around. Sum of moments around some point. Well, we can eliminate both ray and rax if we sum the moments around point a. So that's the choice that we're going to make is we're going to do our sum of moments around point a. So I'll do a sketch here of what the beam looks like. We're trying to take the moments around point a here. So we know that we have our tension, the two components of our tension being applied at point b, and the component of our weight, one component pointing down at point c. So let's again define our positive moments in a clockwise direction. Let's look at each of the pieces here. I'm going to start on the far right, noting that the weight itself is going to make me move or tend to make a moment in the clockwise direction around point a. So we'll take the weight and we'll multiply it by the moment arm. Well, we've established that that moment arm is this long distance all the way to the outside, which I believe we said was l1 plus l2 times the cosine of 30 degrees. Now we consider the moments caused by the tension. Let's look at the tension in the y direction. Here is ty. And we'll notice that that has a moment arm. If we extend the line of action, that moment arm is this shorter distance here that we measured before. We also notice that this is tending to make this spin in a counterclockwise direction. So we're going to have a negative sign. There's the magnitude of my force ty. And we're going to make that go in at a distance, apply that at a moment arm, l1 times cosine of 30 degrees. Notice ty is actually our t sign of 15 degrees. So we'll need to plug that in in just a second. And we're going to have multiple signs and cosines in here. So this last piece is the tension in the x direction, the component in the x direction of the tension, times l1 times the sine of 30 degrees, which is the vertical moment arm up to that x component. And we know that the sum of all of these is going to be equal again to 0. Well, if I go ahead and replace the ty with t sign of 15 degrees and the tx with t cosine of 15 degrees, I will only have one unknown in this equation. And that's the value t. So doing the appropriate algebra, moving everything with the t to one side and then dividing out all the parts that go along with the t, we get something like this. Let's first move the t to one side. We have t times l1 sine of 15 degrees cosine of 30 degrees plus l1 sine of 30 degrees cosine of 15 degrees is equal to w l1 plus l2 cosine of 30 degrees. Now you could either type all this into your calculator or you could recognize one of the things from a trigonometry class that this is one of your adding angle formulas. This is actually the same as the sine of 30 plus 15, or in other words, the sine of 45 degrees. That's one simplification you can make there. Although if you just type each of those pieces into your calculator, you'll end up with the same results. So I now know that my tension, t, is going to be equal to the weight times that total length times the cosine of 30 degrees divided by, here's an l1 we need to divide out, l1, that length times the sine of 45 degrees. And if I plug in all the various pieces there, I'll be able to solve for the tension. A weight of 3 kilonewtons, a total length of 2 meters plus 3 meters times the cosine of 30 degrees over a shorter, I mean, an l1 of 3 meters times the sine of 45 degrees. And that ends up being, well, the cosine of 30 degrees is 0.8660. And the sine of 45 is 0.7071. We'll notice that meters will cancel in this case, giving us an answer in kilonewtons. We can also simplify by noticing that the 3 will cancel out here. So when we do the math, we end up with 6.12 kilonewtons for the tension, the amount of force that the cable is exerting to hold up the weight. Well, now we can go back to our other equations. We recognize from our previous equation that rax was equal to the tension times the cosine of my 15 degrees. Well, I plug that in. The cosine of the 15 degrees is 0.9659. And the tension is the value that we just calculated. So I get my reaction force in the x direction must be 5.91 kilonewtons. Well, let's go back up and take a look, shall we? rAy plus T sine of 15 degrees is equal to 3 kilonewtons. So rAy is equal to 3 kilonewtons minus T sine of 15 degrees. The sine of 15 degrees is 0.2588. And if we solve that, we get an answer of 1.42 kilonewtons. So now we know the reaction forces at the base of the beam as well as the tension holding it up. Notice these numbers, particularly the tension, is significantly larger than the weight itself. Just as a point, I'd like to take a look and point out that you can consider this problem from another perspective. And if we had decided to look at the problem by twisting ourselves, rotating ourselves 30 degrees, moving the entire system down so that the beam itself was oriented along the horizontal axis, we could consider two. We could again consider our reactions. But in case, we've got to be calling these reaction x and reaction y, because they're no longer oriented in the x and y planes. I'm going to call this the reaction parallel to the beam and the reaction perpendicular to the beam. And now I recognize that this angle here is 45 degrees total and that the weight is no longer acting perfectly vertical, but it's acting at a 30 degree angle. If I do that instead, it simplifies part of the problem in the fact that now my moment arms, the only moment arms that are important, are along the beam itself. It's the entire beam of 5 meters and the shorter part of the beam of 3 meters. So I could recreate the equations if I now call this my x direction, or maybe we'll call it x prime and this my y direction, but we'll call it y prime to distinguish from the earlier version. So now that we've established our new basis, our rotated basis, let's go ahead and do static equilibrium in that basis. We'll take the sum of forces in the x prime direction, the sum of forces in the y prime direction, and then the sum of moments around some point. And again, because it's a rotated basis, it doesn't change how we would do some of moments. We'll do the sum of moments about point A again. In this case, the sum of forces in the x direction is going to be equal to our parallel component and then our tension component times the cosine of 45 degrees and our weight component times the cosine of 30 degrees. Notice the tension and the weight are acting to the left, or at least their x components are. For the y prime direction, we have the perpendicular component of the reaction force plus a positive tension times 45 degrees and a negative weight times the sine of 30 degrees. Notice there are substantially more trigonometry with the forces themselves, because the forces are no longer, or at least the weight itself, isn't aligned with my new rotated coordinate system. However, I'm actually going to change the idea. Instead of taking the moments of brown point A, I'm going to make this even simpler and consider the moments around point B, around this point here in the center. If I do so, I could ignore the tension because it's running through point B and has a moment arm of 0. So now if I look at that, I can also ignore our parallel because it is collinear with point B and runs through it. So now I can simply consider our perpendicular applied at a distance of L1 at 3 meters, and that's going to give me a positive moment. And then I will also see that the weight itself, its x component of the weight doesn't apply, but the y component of the weight, W sine of 30 degrees, is also acting in a positive moment direction at the distance of L2. And each of these equations is, again, sums to be equal to 0. If you solve these equations, they are algebraically equivalent to the equations that we solved before. Some aspects of the problem, when viewed from a rotated frame of reference, make it easier to solve. And that is our set of examples for static equilibrium of beams.