WEBVTT 00:00:00.058 --> 00:00:03.969 This is a planar truss: straight bars pinned together at their ends, 00:00:04.052 --> 00:00:06.224 all lying in one plane. 00:00:06.420 --> 00:00:08.659 Here is the one we are going to analyse, 00:00:08.742 --> 00:00:11.704 and it is the only structure you will see today. 00:00:12.245 --> 00:00:14.022 There are four joints. 00:00:14.218 --> 00:00:18.977 A at the left end of the bottom chord, B in the middle of it, C at the right end, 00:00:19.060 --> 00:00:20.617 and D at the peak. 00:00:21.669 --> 00:00:26.752 Each bottom panel is four metres across, and the peak stands three metres above B. 00:00:26.835 --> 00:00:29.005 So each diagonal is five metres, 00:00:29.088 --> 00:00:34.081 and that three four five triangle is the reason every number today comes out whole. 00:00:34.994 --> 00:00:39.418 One load hangs at the peak: twelve kilonewtons, straight down. 00:00:39.614 --> 00:00:42.413 Nothing else is applied anywhere on this frame. 00:00:43.628 --> 00:00:45.173 And the supports. 00:00:45.276 --> 00:00:49.805 At A there is a pin, which can push back both sideways and upward. 00:00:50.001 --> 00:00:53.648 At C there is a roller, which can only push straight up. 00:00:55.300 --> 00:00:59.106 Before any arithmetic, three idealisations do the heavy lifting, 00:00:59.189 --> 00:01:02.916 and they are what make a truss so much easier than a general frame. 00:01:03.829 --> 00:01:07.429 The bars are pinned, so no joint can transmit a moment. 00:01:07.625 --> 00:01:11.689 Loads act only at the joints, never partway along a bar. 00:01:12.141 --> 00:01:15.193 And the weight of each bar is small compared with what it carries, 00:01:15.276 --> 00:01:16.635 so we drop it. 00:01:17.548 --> 00:01:22.843 Put those together and every bar is loaded at exactly two points, its two end pins. 00:01:23.039 --> 00:01:27.928 A body held in equilibrium by forces at two points only can do one thing. 00:01:28.125 --> 00:01:33.570 Those two forces must be equal, opposite, and along the line joining the points. 00:01:34.367 --> 00:01:38.408 So each bar carries force along its own axis and nothing else. 00:01:38.721 --> 00:01:39.951 Bars like that are called 00:01:39.951 --> 00:01:44.062 two force members, and a truss is simply a collection of them. 00:01:45.041 --> 00:01:48.768 Which means there are exactly two things such a bar can be doing. 00:01:48.965 --> 00:01:53.134 Here is a single bar, on its own, with the two joints it connects. 00:01:53.744 --> 00:01:56.506 If the bar is being stretched, it pulls back. 00:01:56.589 --> 00:02:00.049 It drags both of its joints inward, toward each other. 00:02:00.246 --> 00:02:01.614 That is tension, 00:02:01.697 --> 00:02:05.750 and these two green arrows are the forces the bar applies to the joints. 00:02:06.546 --> 00:02:10.051 If instead the bar is being squashed, it shoves back. 00:02:10.134 --> 00:02:13.441 It pushes both joints outward, away from each other. 00:02:13.524 --> 00:02:17.646 That is compression, and the red arrows point the opposite way. 00:02:18.443 --> 00:02:21.485 Those two words are the whole answer we are after. 00:02:21.589 --> 00:02:25.420 By the end of this lecture every one of the five bars will carry a number and one 00:02:25.420 --> 00:02:26.930 of those two letters. 00:02:27.727 --> 00:02:29.481 And here is the promise. 00:02:29.794 --> 00:02:33.382 You will not have to memorise a sign rule to get the letter right. 00:02:33.486 --> 00:02:35.086 A negative answer will mean, 00:02:35.169 --> 00:02:38.734 quite literally, that we drew the arrow the wrong way round. 00:02:39.822 --> 00:02:43.977 Before solving anything, it is worth asking whether it can be solved at all. 00:02:44.060 --> 00:02:47.578 So here is the same frame again, and we are going to count. 00:02:48.375 --> 00:02:50.222 First the unknowns. 00:02:50.418 --> 00:02:55.316 There are five bars, and each carries one unknown force along its own axis. 00:02:55.399 --> 00:02:58.999 So that is five numbers we do not know yet, one for each bar. 00:02:59.795 --> 00:03:01.164 Now the supports. 00:03:01.247 --> 00:03:05.172 Rub each one out and replace it by the forces it is able to exert. 00:03:05.484 --> 00:03:10.581 The pin at A can push sideways and it can push up, so that is two more unknowns. 00:03:10.778 --> 00:03:14.366 The roller at C can only push up, so that is one more. 00:03:15.279 --> 00:03:18.391 Five bar forces and three reaction components. 00:03:18.588 --> 00:03:23.302 Eight unknown numbers in total, and that is the left hand side of the comparison. 00:03:24.517 --> 00:03:28.046 Now the equations, and here is where a truss is special. 00:03:28.243 --> 00:03:32.714 At a joint, every force in the picture passes through one single point. 00:03:32.910 --> 00:03:35.405 Forces through a point have no moment about it, 00:03:35.488 --> 00:03:38.066 so there is no moment equation to write. 00:03:38.770 --> 00:03:42.543 That leaves two equations at each joint, and only two. 00:03:42.740 --> 00:03:47.361 The horizontal forces sum to zero, and the vertical forces sum to zero. 00:03:47.558 --> 00:03:51.402 Four joints, two equations apiece, is eight equations. 00:03:52.198 --> 00:03:54.777 Eight unknowns and eight equations. 00:03:54.973 --> 00:03:59.211 They match exactly, and that is what statically determinate means. 00:03:59.408 --> 00:04:03.344 Equilibrium on its own is enough to find every force, with nothing left 00:04:03.344 --> 00:04:05.063 over and nothing missing. 00:04:06.459 --> 00:04:09.360 It is worth knowing what the other two answers would have meant, 00:04:09.443 --> 00:04:11.325 because you will meet both. 00:04:12.028 --> 00:04:14.175 If the unknowns fall short of the equations, 00:04:14.258 --> 00:04:16.488 there are too few bars to hold the shape. 00:04:16.800 --> 00:04:18.424 The frame is a mechanism. 00:04:18.507 --> 00:04:21.050 It folds up rather than carrying anything. 00:04:21.847 --> 00:04:26.443 If the unknowns outnumber the equations, the frame is statically indeterminate. 00:04:26.526 --> 00:04:30.543 It stands perfectly well, but equilibrium alone will not tell you how the load 00:04:30.543 --> 00:04:34.897 shares itself out, and you would need to bring in how much each bar stretches. 00:04:35.694 --> 00:04:38.364 Ours is the middle case, so the method of joints will run 00:04:38.364 --> 00:04:40.049 all the way through to the end. 00:04:40.245 --> 00:04:43.822 But it has to start somewhere, and joints are not where it starts. 00:04:44.925 --> 00:04:49.744 The method of joints needs a place to start, and joints are the wrong place. 00:04:50.359 --> 00:04:54.444 Every one of them has too many unknowns until we know what the ground is doing. 00:04:54.527 --> 00:04:59.218 So step one is always the same: forget that this is a truss at all. 00:04:59.829 --> 00:05:03.206 Treat the whole thing as one rigid body, floating free, 00:05:03.289 --> 00:05:07.573 with only the applied load and the three support reactions acting on it. 00:05:07.886 --> 00:05:10.476 What happens inside does not matter yet. 00:05:11.272 --> 00:05:15.917 For a rigid body in a plane there are exactly three equations available. 00:05:16.114 --> 00:05:20.744 The horizontal forces sum to zero, the vertical forces sum to zero, 00:05:20.827 --> 00:05:24.056 and the moments about any point you like sum to zero. 00:05:24.252 --> 00:05:26.958 Three equations, three unknowns. 00:05:28.010 --> 00:05:31.146 Take moments about A, and take them first. 00:05:31.458 --> 00:05:33.735 That choice is not an accident. 00:05:33.931 --> 00:05:37.797 Both of the reactions at A pass straight through A, so neither of them 00:05:37.797 --> 00:05:42.488 has any moment about it, and both drop out of the equation before we write it. 00:05:43.285 --> 00:05:47.232 That leaves two terms. The twelve kilonewton load acts four metres 00:05:47.232 --> 00:05:50.844 to the right of A and turns the truss clockwise about it. 00:05:51.156 --> 00:05:55.848 The reaction at C acts eight metres to the right and turns it the other way. 00:05:56.644 --> 00:05:58.503 Set the sum to zero. 00:05:58.699 --> 00:06:03.274 Eight times C y, minus twelve times four, equals nothing. 00:06:04.070 --> 00:06:06.811 Twelve times four is forty eight, 00:06:07.008 --> 00:06:10.422 so C y is six kilonewtons, pushing upward. 00:06:11.636 --> 00:06:17.163 Now vertical forces. Six kilonewtons up at C, twelve down at the peak, and A y, 00:06:17.163 --> 00:06:19.021 whatever it turns out to be. 00:06:19.934 --> 00:06:24.483 So A y is six kilonewtons upward as well. And that ought to feel right. 00:06:24.566 --> 00:06:27.537 The load sits exactly halfway between the two supports, 00:06:27.620 --> 00:06:29.896 so of course they share it evenly. 00:06:30.321 --> 00:06:32.073 Horizontal forces last. 00:06:32.156 --> 00:06:35.521 The load points straight down and the roller can only push straight up, 00:06:35.604 --> 00:06:38.586 so there is nothing in the entire picture pushing sideways. 00:06:38.669 --> 00:06:40.794 A x has nothing to balance. 00:06:41.138 --> 00:06:45.713 Which makes it zero. So we can rub that arrow out and never think about it again. 00:06:47.110 --> 00:06:50.373 Those two sixes are the doorway into the truss. 00:06:50.570 --> 00:06:54.692 Up until now, joint A had three things we did not know at it. 00:06:54.796 --> 00:06:56.213 Now it has two. 00:06:56.696 --> 00:06:58.357 And two is the magic number, 00:06:58.357 --> 00:07:02.003 because two is exactly how many equations a joint gives us. 00:07:02.199 --> 00:07:05.727 So from here the whole analysis is a walk from joint to joint, 00:07:05.810 --> 00:07:08.156 spending two equations at each one. 00:07:09.253 --> 00:07:12.874 Now we walk the joints, and the rule for choosing one is simple. 00:07:12.957 --> 00:07:16.371 Go where no more than two bar forces are unknown. 00:07:16.474 --> 00:07:20.248 Joint A qualifies, because only two bars meet there. 00:07:20.732 --> 00:07:24.564 So cut joint A out of the structure and look at it on its own. 00:07:24.760 --> 00:07:29.684 The reaction we just found is here, six kilonewtons pushing the joint upward. 00:07:30.387 --> 00:07:31.746 The two bars that were attached 00:07:31.746 --> 00:07:36.135 to it are gone, so we have to put back the force each of them was applying. 00:07:36.332 --> 00:07:37.086 Here is the rule 00:07:37.086 --> 00:07:41.696 that makes all the signs work, and it is the only convention in this lecture. 00:07:42.608 --> 00:07:46.104 Draw every unknown bar force pulling away from the joint. 00:07:46.416 --> 00:07:48.179 Both arrows point outward, 00:07:48.262 --> 00:07:52.478 straight along their own bars, as though every single bar were in tension. 00:07:53.274 --> 00:07:55.260 We are not claiming they are. 00:07:55.457 --> 00:07:56.955 We are only choosing a direction 00:07:56.955 --> 00:08:01.030 so that the algebra has something to be positive or negative about. 00:08:01.482 --> 00:08:05.071 If the answer comes back positive the bar really is pulling. 00:08:05.174 --> 00:08:08.180 If it comes back negative the true arrow points the other way, 00:08:08.263 --> 00:08:09.982 and the bar is pushing. 00:08:10.686 --> 00:08:13.296 One piece of geometry before we resolve. 00:08:13.379 --> 00:08:18.219 Bar A D rises three for every four across, and its length is five. 00:08:18.302 --> 00:08:23.737 So the angle theta at A has a sine of three fifths and a cosine of four fifths. 00:08:25.133 --> 00:08:29.198 Take the vertical equation first, and take it first for a reason. 00:08:29.812 --> 00:08:34.109 F A B is horizontal, so it has no vertical part at all. 00:08:34.305 --> 00:08:38.776 It contributes nothing to this sum, which leaves only one unknown in it. 00:08:40.176 --> 00:08:42.740 Six kilonewtons upward from the reaction, 00:08:42.823 --> 00:08:46.806 plus the vertical part of F A D, must come to zero. 00:08:47.421 --> 00:08:53.192 That vertical part is F A D times the sine of theta, so it is three fifths of it. 00:08:53.895 --> 00:09:00.177 Three fifths of F A D has to cancel six, so F A D is minus ten kilonewtons. 00:09:00.974 --> 00:09:02.888 And there is the negative. 00:09:02.971 --> 00:09:07.453 It is not a bookkeeping nuisance, it is the answer telling us something. 00:09:07.557 --> 00:09:10.088 The arrow we drew is backwards. 00:09:10.699 --> 00:09:13.983 The bar is not pulling joint A up along the diagonal. 00:09:14.066 --> 00:09:17.364 It is shoving it down along the diagonal, like this. 00:09:17.561 --> 00:09:20.034 Ten kilonewtons of compression. 00:09:21.431 --> 00:09:26.830 With F A D known, the horizontal equation has just one unknown left in it. 00:09:27.027 --> 00:09:30.673 So we spend the second of our two equations at this joint. 00:09:31.470 --> 00:09:35.731 F A B is entirely horizontal, so all of it counts. 00:09:36.044 --> 00:09:40.628 The horizontal part of F A D is F A D times the cosine of theta, 00:09:40.711 --> 00:09:42.639 which is four fifths of it. 00:09:43.552 --> 00:09:47.999 Four fifths of minus ten is minus eight, pointing to the left. 00:09:48.196 --> 00:09:52.330 So F A B must be plus eight kilonewtons to balance it. 00:09:53.242 --> 00:09:56.816 Positive this time, so the arrow we drew was already right. 00:09:56.899 --> 00:10:00.395 Bar A B really does pull joint A to the right. 00:10:00.591 --> 00:10:02.786 Eight kilonewtons of tension. 00:10:03.490 --> 00:10:06.298 And notice that the two results explain each other. 00:10:06.381 --> 00:10:08.225 The diagonal is in compression, 00:10:08.308 --> 00:10:11.618 so it is shoving the bottom corner of the truss outward. 00:10:12.070 --> 00:10:16.576 The bottom chord is what stops that happening, so of course it is stretched. 00:10:17.972 --> 00:10:19.689 Step back to the whole truss, 00:10:19.772 --> 00:10:24.161 with the reactions we found and the three bar forces we have not touched yet. 00:10:25.074 --> 00:10:28.450 The left diagonal is in compression at ten kilonewtons, 00:10:28.533 --> 00:10:31.111 and from here on it is drawn in red. 00:10:31.424 --> 00:10:35.999 The left half of the bottom chord is in tension at eight, and that one is green. 00:10:36.703 --> 00:10:39.513 So two bars are done, and three remain. 00:10:39.710 --> 00:10:42.994 And the next joint we visit will not need any algebra at all, 00:10:43.077 --> 00:10:47.107 because the answer there can be had just by looking carefully at the picture. 00:10:48.206 --> 00:10:51.130 Joint B next. Three bars meet here. 00:10:51.213 --> 00:10:54.752 A B coming in from the left, B C going out to the right, 00:10:54.835 --> 00:10:59.260 and the vertical B D going up. And nothing at all is applied at this joint. 00:11:00.056 --> 00:11:03.119 Before writing a single equation, just look at it. 00:11:03.202 --> 00:11:07.394 Two of those three bars lie along one straight line, the bottom chord. 00:11:07.591 --> 00:11:10.540 Both of their forces are purely horizontal. 00:11:11.337 --> 00:11:13.252 So the vertical bar is the only thing 00:11:13.252 --> 00:11:16.446 at this joint with any vertical component whatsoever. 00:11:16.550 --> 00:11:19.186 There is nothing for it to balance against. 00:11:19.890 --> 00:11:25.148 Which means vertical equilibrium at B gives F B D equals zero, immediately, 00:11:25.231 --> 00:11:27.147 with no arithmetic at all. 00:11:27.344 --> 00:11:28.796 Away it goes. 00:11:29.848 --> 00:11:34.237 That is a zero force member, and the pattern is worth committing to memory. 00:11:34.434 --> 00:11:36.417 Three bars at an unloaded joint, 00:11:36.500 --> 00:11:40.494 two of them in a straight line, and the odd one out carries nothing. 00:11:41.407 --> 00:11:43.611 It is not a useless bar, though. 00:11:43.694 --> 00:11:46.073 It braces the bottom chord against buckling, 00:11:46.156 --> 00:11:49.918 and the instant anything is hung at B it starts carrying load. 00:11:50.231 --> 00:11:53.935 Under this particular loading it simply happens to carry none. 00:11:54.360 --> 00:11:56.311 Then horizontally. 00:11:56.508 --> 00:11:58.712 With nothing pushing sideways at B, 00:11:58.795 --> 00:12:01.338 the two chord forces have to be equal and opposite, 00:12:01.338 --> 00:12:07.214 so F B C is the same eight kilonewtons of tension we already found in A B. 00:12:08.355 --> 00:12:11.223 Joint C now, and it will look familiar. 00:12:11.327 --> 00:12:14.253 The roller pushes up with six kilonewtons. 00:12:14.357 --> 00:12:17.850 Bar C B pulls the joint to the left with the eight we just found. 00:12:17.933 --> 00:12:21.707 And the diagonal C D is the last unknown in the whole structure. 00:12:22.411 --> 00:12:25.416 Same convention, drawn pulling away from the joint. 00:12:25.499 --> 00:12:31.014 Same geometry too, because this diagonal is also a three four five, just mirrored. 00:12:31.811 --> 00:12:37.861 Vertical equilibrium. Six upward, plus three fifths of F C D, equals zero. 00:12:37.964 --> 00:12:41.506 So F C D is minus ten kilonewtons. 00:12:42.302 --> 00:12:45.577 Negative again, so once again the arrow flips. 00:12:45.890 --> 00:12:48.559 The right diagonal is pushing on joint C, 00:12:48.642 --> 00:12:52.694 ten kilonewtons of compression, exactly mirroring the left one. 00:12:53.398 --> 00:12:55.834 And now something rather nice happens. 00:12:55.917 --> 00:12:57.671 Every bar force is already known, 00:12:57.671 --> 00:13:01.770 so the horizontal equation at C has no unknowns left in it. 00:13:01.966 --> 00:13:04.730 It is not a tool any more, it is a test. 00:13:05.527 --> 00:13:09.811 Eight kilonewtons pulling the joint left from the chord, and four fifths 00:13:09.811 --> 00:13:14.188 of minus ten from the diagonal, which is a push of eight to the right. 00:13:14.501 --> 00:13:18.693 They cancel exactly, and that is the arithmetic checking itself. 00:13:19.996 --> 00:13:22.502 Here is the whole structure with everything on it. 00:13:22.585 --> 00:13:26.650 The bottom chord in green, in tension across both panels. 00:13:26.846 --> 00:13:29.820 Both diagonals in red, in compression. 00:13:30.016 --> 00:13:32.792 And the vertical, carrying nothing. 00:13:33.588 --> 00:13:35.725 We used joints A, B and C, 00:13:35.725 --> 00:13:40.462 which is three joints and six equations, and we had eight equations available. 00:13:40.659 --> 00:13:45.106 So joint D is left over, and that makes it a free check on all of the work. 00:13:45.810 --> 00:13:47.886 Both diagonals are in compression, 00:13:47.969 --> 00:13:52.626 so both of them push on D, outward and away from their far ends. 00:13:52.822 --> 00:13:57.360 Horizontally, the left one shoves D to the right by four fifths of ten, 00:13:57.443 --> 00:14:00.761 and the right one shoves it left by exactly the same eight. 00:14:00.844 --> 00:14:02.134 They cancel. 00:14:02.838 --> 00:14:06.458 Vertically, each diagonal lifts D by three fifths of ten, 00:14:06.541 --> 00:14:08.817 which is six kilonewtons apiece. 00:14:09.014 --> 00:14:14.077 Twelve upward in total, carrying precisely the twelve kilonewtons hanging there. 00:14:14.273 --> 00:14:17.583 The vertical bar contributes nothing, as we found. 00:14:18.980 --> 00:14:22.301 So here is the finished truss, gathered up. 00:14:22.405 --> 00:14:26.376 The bottom chord pulls at eight kilonewtons in both panels. 00:14:26.573 --> 00:14:28.803 Both diagonals push at ten. 00:14:28.999 --> 00:14:31.055 The vertical carries nothing. 00:14:31.852 --> 00:14:33.151 And look at the pattern, 00:14:33.234 --> 00:14:37.019 because it is the pattern of almost every simple truss you will meet. 00:14:37.216 --> 00:14:40.119 Imagine this one sagging under its load. 00:14:40.316 --> 00:14:42.755 The bottom stretches, so it pulls. 00:14:42.951 --> 00:14:46.319 The bars nearer the top squash, so they push. 00:14:47.232 --> 00:14:49.253 That is the whole method. 00:14:49.705 --> 00:14:52.980 Find the reactions from the truss as one body. 00:14:53.176 --> 00:14:55.998 Then walk from joint to joint, never taking 00:14:55.998 --> 00:15:00.503 on more than two unknowns at a time, spending two equations at each stop. 00:15:01.207 --> 00:15:06.479 And at every one of those stops, draw the unknown arrow pulling away from the joint. 00:15:06.675 --> 00:15:09.973 If the number comes back positive, the bar pulls. 00:15:10.286 --> 00:15:13.305 If it comes back negative, the bar pushes. 00:15:13.502 --> 00:15:18.611 The sign is not bookkeeping. It is telling you which way the arrow really points.