Planar Truss Analysis: Method of Joints

This is a planar truss: straight bars pinned together at their ends, all lying in one plane. Here is the one we are going to analyse, and it is the only structure you will see today. There are four joints. A at the left end of the bottom chord, B in the middle of it, C at the right end, and D at the peak. Each bottom panel is four metres across, and the peak stands three metres above B. So each diagonal is five metres, and that three four five triangle is the reason every number today comes out whole. One load hangs at the peak: twelve kilonewtons, straight down. Nothing else is applied anywhere on this frame. And the supports. At A there is a pin, which can push back both sideways and upward. At C there is a roller, which can only push straight up. Before any arithmetic, three idealisations do the heavy lifting, and they are what make a truss so much easier than a general frame. The bars are pinned, so no joint can transmit a moment. Loads act only at the joints, never partway along a bar. And the weight of each bar is small compared with what it carries, so we drop it. Put those together and every bar is loaded at exactly two points, its two end pins. A body held in equilibrium by forces at two points only can do one thing. Those two forces must be equal, opposite, and along the line joining the points. So each bar carries force along its own axis and nothing else. Bars like that are called two force members, and a truss is simply a collection of them. Which means there are exactly two things such a bar can be doing. Here is a single bar, on its own, with the two joints it connects. If the bar is being stretched, it pulls back. It drags both of its joints inward, toward each other. That is tension, and these two green arrows are the forces the bar applies to the joints. If instead the bar is being squashed, it shoves back. It pushes both joints outward, away from each other. That is compression, and the red arrows point the opposite way. Those two words are the whole answer we are after. By the end of this lecture every one of the five bars will carry a number and one of those two letters. And here is the promise. You will not have to memorise a sign rule to get the letter right. A negative answer will mean, quite literally, that we drew the arrow the wrong way round.

Before solving anything, it is worth asking whether it can be solved at all. So here is the same frame again, and we are going to count. First the unknowns. There are five bars, and each carries one unknown force along its own axis. So that is five numbers we do not know yet, one for each bar. Now the supports. Rub each one out and replace it by the forces it is able to exert. The pin at A can push sideways and it can push up, so that is two more unknowns. The roller at C can only push up, so that is one more. Five bar forces and three reaction components. Eight unknown numbers in total, and that is the left hand side of the comparison. Now the equations, and here is where a truss is special. At a joint, every force in the picture passes through one single point. Forces through a point have no moment about it, so there is no moment equation to write. That leaves two equations at each joint, and only two. The horizontal forces sum to zero, and the vertical forces sum to zero. Four joints, two equations apiece, is eight equations. Eight unknowns and eight equations. They match exactly, and that is what statically determinate means. Equilibrium on its own is enough to find every force, with nothing left over and nothing missing. It is worth knowing what the other two answers would have meant, because you will meet both. If the unknowns fall short of the equations, there are too few bars to hold the shape. The frame is a mechanism. It folds up rather than carrying anything. If the unknowns outnumber the equations, the frame is statically indeterminate. It stands perfectly well, but equilibrium alone will not tell you how the load shares itself out, and you would need to bring in how much each bar stretches. Ours is the middle case, so the method of joints will run all the way through to the end. But it has to start somewhere, and joints are not where it starts.

The method of joints needs a place to start, and joints are the wrong place. Every one of them has too many unknowns until we know what the ground is doing. So step one is always the same: forget that this is a truss at all. Treat the whole thing as one rigid body, floating free, with only the applied load and the three support reactions acting on it. What happens inside does not matter yet. For a rigid body in a plane there are exactly three equations available. The horizontal forces sum to zero, the vertical forces sum to zero, and the moments about any point you like sum to zero. Three equations, three unknowns. Take moments about A, and take them first. That choice is not an accident. Both of the reactions at A pass straight through A, so neither of them has any moment about it, and both drop out of the equation before we write it. That leaves two terms. The twelve kilonewton load acts four metres to the right of A and turns the truss clockwise about it. The reaction at C acts eight metres to the right and turns it the other way. Set the sum to zero. Eight times C y, minus twelve times four, equals nothing. Twelve times four is forty eight, so C y is six kilonewtons, pushing upward. Now vertical forces. Six kilonewtons up at C, twelve down at the peak, and A y, whatever it turns out to be. So A y is six kilonewtons upward as well. And that ought to feel right. The load sits exactly halfway between the two supports, so of course they share it evenly. Horizontal forces last. The load points straight down and the roller can only push straight up, so there is nothing in the entire picture pushing sideways. A x has nothing to balance. Which makes it zero. So we can rub that arrow out and never think about it again. Those two sixes are the doorway into the truss. Up until now, joint A had three things we did not know at it. Now it has two. And two is the magic number, because two is exactly how many equations a joint gives us. So from here the whole analysis is a walk from joint to joint, spending two equations at each one.

Now we walk the joints, and the rule for choosing one is simple. Go where no more than two bar forces are unknown. Joint A qualifies, because only two bars meet there. So cut joint A out of the structure and look at it on its own. The reaction we just found is here, six kilonewtons pushing the joint upward. The two bars that were attached to it are gone, so we have to put back the force each of them was applying. Here is the rule that makes all the signs work, and it is the only convention in this lecture. Draw every unknown bar force pulling away from the joint. Both arrows point outward, straight along their own bars, as though every single bar were in tension. We are not claiming they are. We are only choosing a direction so that the algebra has something to be positive or negative about. If the answer comes back positive the bar really is pulling. If it comes back negative the true arrow points the other way, and the bar is pushing. One piece of geometry before we resolve. Bar A D rises three for every four across, and its length is five. So the angle theta at A has a sine of three fifths and a cosine of four fifths. Take the vertical equation first, and take it first for a reason. F A B is horizontal, so it has no vertical part at all. It contributes nothing to this sum, which leaves only one unknown in it. Six kilonewtons upward from the reaction, plus the vertical part of F A D, must come to zero. That vertical part is F A D times the sine of theta, so it is three fifths of it. Three fifths of F A D has to cancel six, so F A D is minus ten kilonewtons. And there is the negative. It is not a bookkeeping nuisance, it is the answer telling us something. The arrow we drew is backwards. The bar is not pulling joint A up along the diagonal. It is shoving it down along the diagonal, like this. Ten kilonewtons of compression. With F A D known, the horizontal equation has just one unknown left in it. So we spend the second of our two equations at this joint. F A B is entirely horizontal, so all of it counts. The horizontal part of F A D is F A D times the cosine of theta, which is four fifths of it. Four fifths of minus ten is minus eight, pointing to the left. So F A B must be plus eight kilonewtons to balance it. Positive this time, so the arrow we drew was already right. Bar A B really does pull joint A to the right. Eight kilonewtons of tension. And notice that the two results explain each other. The diagonal is in compression, so it is shoving the bottom corner of the truss outward. The bottom chord is what stops that happening, so of course it is stretched. Step back to the whole truss, with the reactions we found and the three bar forces we have not touched yet. The left diagonal is in compression at ten kilonewtons, and from here on it is drawn in red. The left half of the bottom chord is in tension at eight, and that one is green. So two bars are done, and three remain. And the next joint we visit will not need any algebra at all, because the answer there can be had just by looking carefully at the picture.

Joint B next. Three bars meet here. A B coming in from the left, B C going out to the right, and the vertical B D going up. And nothing at all is applied at this joint. Before writing a single equation, just look at it. Two of those three bars lie along one straight line, the bottom chord. Both of their forces are purely horizontal. So the vertical bar is the only thing at this joint with any vertical component whatsoever. There is nothing for it to balance against. Which means vertical equilibrium at B gives F B D equals zero, immediately, with no arithmetic at all. Away it goes. That is a zero force member, and the pattern is worth committing to memory. Three bars at an unloaded joint, two of them in a straight line, and the odd one out carries nothing. It is not a useless bar, though. It braces the bottom chord against buckling, and the instant anything is hung at B it starts carrying load. Under this particular loading it simply happens to carry none. Then horizontally. With nothing pushing sideways at B, the two chord forces have to be equal and opposite, so F B C is the same eight kilonewtons of tension we already found in A B. Joint C now, and it will look familiar. The roller pushes up with six kilonewtons. Bar C B pulls the joint to the left with the eight we just found. And the diagonal C D is the last unknown in the whole structure. Same convention, drawn pulling away from the joint. Same geometry too, because this diagonal is also a three four five, just mirrored. Vertical equilibrium. Six upward, plus three fifths of F C D, equals zero. So F C D is minus ten kilonewtons. Negative again, so once again the arrow flips. The right diagonal is pushing on joint C, ten kilonewtons of compression, exactly mirroring the left one. And now something rather nice happens. Every bar force is already known, so the horizontal equation at C has no unknowns left in it. It is not a tool any more, it is a test. Eight kilonewtons pulling the joint left from the chord, and four fifths of minus ten from the diagonal, which is a push of eight to the right. They cancel exactly, and that is the arithmetic checking itself. Here is the whole structure with everything on it. The bottom chord in green, in tension across both panels. Both diagonals in red, in compression. And the vertical, carrying nothing. We used joints A, B and C, which is three joints and six equations, and we had eight equations available. So joint D is left over, and that makes it a free check on all of the work. Both diagonals are in compression, so both of them push on D, outward and away from their far ends. Horizontally, the left one shoves D to the right by four fifths of ten, and the right one shoves it left by exactly the same eight. They cancel. Vertically, each diagonal lifts D by three fifths of ten, which is six kilonewtons apiece. Twelve upward in total, carrying precisely the twelve kilonewtons hanging there. The vertical bar contributes nothing, as we found. So here is the finished truss, gathered up. The bottom chord pulls at eight kilonewtons in both panels. Both diagonals push at ten. The vertical carries nothing. And look at the pattern, because it is the pattern of almost every simple truss you will meet. Imagine this one sagging under its load. The bottom stretches, so it pulls. The bars nearer the top squash, so they push. That is the whole method. Find the reactions from the truss as one body. Then walk from joint to joint, never taking on more than two unknowns at a time, spending two equations at each stop. And at every one of those stops, draw the unknown arrow pulling away from the joint. If the number comes back positive, the bar pulls. If it comes back negative, the bar pushes. The sign is not bookkeeping. It is telling you which way the arrow really points.