{"version":1,"lectureId":"01M14TYF4MA5MTVE5SWD76QWSG","attempt":0,"publication":{"slug":"planar-truss-method-of-joints","title":"Planar Truss Analysis: Method of Joints","subject":"engineering","summary":"One planar truss, carried from the first sketch to the finished answer. We set up a two panel truss with a single load at its apex, count the unknowns against the available equations to show that it is statically determinate, find the support reactions from whole body equilibrium, and then walk it joint by joint. At every joint we draw each unknown bar force pulling away from the joint, spend the two equilibrium equations available there, and let the sign of the answer say whether the bar pulls or pushes, so that tension and compression stay physical rather than becoming bookkeeping. A zero force member is spotted by inspection on the way through, and the completed truss is checked at the one joint the analysis never needed.","metaDescription":"Analyse one planar truss by the method of joints: support reactions, joint equilibrium, tension and compression signs, zero force members.","transcript":"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.","watch":{"version":1,"scenes":[{"title":"A Truss, and What Its Bars Can Do","start":0,"end":160.9920833333333,"objects":{"bottom_left":"a Line [blue] drawn in truss (end=(4.0, 0.0))","bottom_right":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","c_bar":"a Line [blue] drawn in compression_case (start=(1.6, 1.75), end=(4.0, 1.75))","c_left_arrow":"a Vector [red] drawn in compression_case (start=(1.6, 1.3), end=(0.65, 1.3))","c_left_joint":"a Point [text] drawn in compression_case (location=(1.6, 1.75))","c_left_leader":"a Line [gray] drawn in compression_case (start=(1.6, 1.7), end=(1.6, 1.32), dashed=True)","c_right_arrow":"a Vector [red] drawn in compression_case (start=(4.0, 1.3), end=(4.95, 1.3))","c_right_joint":"a Point [text] drawn in compression_case (location=(4.0, 1.75))","c_right_leader":"a Line [gray] drawn in compression_case (start=(4.0, 1.7), end=(4.0, 1.32), dashed=True)","compression_caption":"a Text [text] that says \"Compression: the bar pushes both joints outward.\"","compression_case":"a Figure (x_range=(0.0, 5.6), y_range=(0.0, 2.8), aspect=(5.6, 2.8))","dim_left":"a Math [gray] that says \"$5 thin upright(\"m\")$\" drawn in truss","dim_right":"a Math [gray] that says \"$5 thin upright(\"m\")$\" drawn in truss","dim_vertical":"a Math [gray] that says \"$3 thin upright(\"m\")$\" drawn in truss","head_ideal":"a Heading that says \"Three Idealisations\"","head_tc":"a Heading that says \"What a Bar Can Do\"","head_truss":"a Heading that says \"The Structure We Will Analyse\"","idealizations":"a Block [text] that says \"The bars are pinned at their ends, so no joint can carry a moment. 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Here is the one we are going to analyse, and it is the only structure you will see today.","live":[],"does":[[0,"head_truss is shown on the screen, written out."],[0,"truss is shown on the screen, written out."],[2.09,"bottom_left is shown on the screen, written out."],[2.3899999999999997,"bottom_right is shown on the screen, written out."],[2.69,"left_diagonal is shown on the screen, written out."],[2.9899999999999998,"right_diagonal is shown on the screen, written out."],[3.29,"vertical is shown on the screen, written out."]]},{"start":11.6645,"say":"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.","live":["truss","head_truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical"],"does":[[13.603,"joint_a is shown on the screen, written out."],[15.912999999999998,"joint_b is shown on the screen, written out."],[17.375999999999998,"joint_c is shown on the screen, written out."],[19.396,"joint_d is shown on the screen, written out."]]},{"start":20.750500000000002,"say":"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.","live":["truss","head_truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d"],"does":[[22.225,"span_left is shown on the screen, written out."],[22.225,"span_right is shown on the screen, written out."],[24.489000000000004,"dim_vertical is shown on the screen, written out."],[27.821000000000005,"dim_left is shown on the screen, written out."],[27.821000000000005,"dim_right is shown on the screen, written out."]]},{"start":34.331,"say":"One load hangs at the peak: twelve kilonewtons, straight down. Nothing else is applied anywhere on this frame.","live":["truss","head_truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","span_left","span_right","dim_vertical","dim_left","dim_right"],"does":[[36.42100000000001,"load is shown on the screen, written out."]]},{"start":42.4305,"say":"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.","live":["truss","head_truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","span_left","span_right","dim_vertical","dim_left","dim_right","load"],"does":[[45.635,"pin_plate is shown on the screen, written out."],[45.635,"pin_ground is shown on the screen, written out."],[50.592000000000006,"roller_plate is shown on the screen, written out."],[50.592000000000006,"roller_left is shown on the screen, written out."],[50.592000000000006,"roller_right is shown on the screen, written out."],[50.592000000000006,"roller_ground is shown on the screen, written out."],[52.926,"truss moves to a new place on the board."],[52.926,"head_truss is hidden from the screen — left the board."]]},{"start":54.126000000000005,"say":"Before any arithmetic, three idealisations do the heavy lifting, and they are what make a truss so much easier than a general frame.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","span_left","span_right","dim_vertical","dim_left","dim_right","load","pin_plate","pin_ground","roller_plate","roller_left","roller_right","roller_ground"],"does":[[54.126000000000005,"head_ideal is shown on the screen, written out."],[55.995000000000005,"idealizations is shown on the screen, written out."]]},{"start":63.0735,"say":"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.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","span_left","span_right","dim_vertical","dim_left","dim_right","load","pin_plate","pin_ground","roller_plate","roller_left","roller_right","roller_ground","idealizations","head_ideal"],"does":[[64.06,"idealizations (the \"pinned\" part) is emphasized."],[67.96100000000001,"idealizations (the \"only at the joints\" part) is emphasized."],[67.96100000000001,"idealizations (the \"pinned\" part) is no longer emphasized."],[74.54400000000001,"idealizations (the \"neglect\" part) is emphasized."],[74.54400000000001,"idealizations (the \"only at the joints\" part) is no longer emphasized."]]},{"start":76.038,"say":"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.","live":null,"does":[[76.038,"idealizations (the \"neglect\" part) is no longer emphasized."],[77.628,"left_diagonal is indicated — a transient flash."]]},{"start":93.4145,"say":"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.","live":null,"does":[[95.94600000000001,"vertical is indicated — a transient flash."],[102.96950000000001,"head_ideal is hidden from the screen — left the board."],[102.96950000000001,"idealizations is hidden from the screen — left the board."],[102.96950000000001,"truss is hidden from the screen — left the board."],[102.96950000000001,"bottom_left is hidden from the screen — truss left the board."],[102.96950000000001,"bottom_right is hidden from the screen — truss left the board."],[102.96950000000001,"left_diagonal is hidden from the screen — truss left the board."],[102.96950000000001,"right_diagonal is hidden from the screen — truss left the board."],[102.96950000000001,"vertical is hidden from the screen — truss left the board."],[102.96950000000001,"joint_a is hidden from the screen — truss left the board."],[102.96950000000001,"joint_b is hidden from the screen — truss left the board."],[102.96950000000001,"joint_c is hidden from the screen — truss left the board."],[102.96950000000001,"joint_d is hidden from the screen — truss left the board."],[102.96950000000001,"span_left is hidden from the screen — truss left the board."],[102.96950000000001,"span_right is hidden from the screen — truss left the board."],[102.96950000000001,"dim_vertical is hidden from the screen — truss left the board."],[102.96950000000001,"dim_left is hidden from the screen — truss left the board."],[102.96950000000001,"dim_right is hidden from the screen — truss left the board."],[102.96950000000001,"load is hidden from the screen — truss left the board."],[102.96950000000001,"pin_plate is hidden from the screen — truss left the board."],[102.96950000000001,"pin_ground is hidden from the screen — truss left the board."],[102.96950000000001,"roller_plate is hidden from the screen — truss left the board."],[102.96950000000001,"roller_left is hidden from the screen — truss left the board."],[102.96950000000001,"roller_right is hidden from the screen — truss left the board."],[102.96950000000001,"roller_ground is hidden from the screen — truss left the board."]]},{"start":104.1695,"say":"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.","live":[],"does":[[104.1695,"head_tc is shown on the screen, written out."],[108.628,"tension_case is shown on the screen, written out."],[108.628,"t_bar is shown on the screen, written out."],[108.628,"t_left_joint is shown on the screen, written out."],[108.628,"t_right_joint is shown on the screen, written out."],[110.90299999999999,"tension_case moves to a new place on the board."],[110.90299999999999,"compression_case is shown on the screen, written out."],[110.90299999999999,"c_bar is shown on the screen, written out."],[110.90299999999999,"c_left_joint is shown on the screen, written out."],[110.90299999999999,"c_right_joint is shown on the screen, written out."]]},{"start":112.8385,"say":"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.","live":["tension_case","compression_case","head_tc","t_bar","t_left_joint","t_right_joint","c_bar","c_left_joint","c_right_joint"],"does":[[116.438,"t_left_leader is shown on the screen, written out."],[116.438,"t_right_leader is shown on the screen, written out."],[120.327,"tension_caption is shown on the screen, written out."],[121.964,"t_left_arrow is shown on the screen, written out."],[121.964,"t_right_arrow is shown on the screen, written out."]]},{"start":125.9425,"say":"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.","live":["tension_case","tension_caption","compression_case","head_tc","t_bar","t_left_joint","t_right_joint","c_bar","c_left_joint","c_right_joint","t_left_leader","t_right_leader","t_left_arrow","t_right_arrow"],"does":[[129.077,"c_left_leader is shown on the screen, written out."],[129.077,"c_right_leader is shown on the screen, written out."],[134.778,"compression_caption is shown on the screen, written out."],[136.055,"c_left_arrow is shown on the screen, written out."],[136.055,"c_right_arrow is shown on the screen, written out."]]},{"start":138.91899999999998,"say":"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.","live":["tension_case","tension_caption","compression_case","compression_caption","head_tc","t_bar","t_left_joint","t_right_joint","c_bar","c_left_joint","c_right_joint","t_left_leader","t_right_leader","t_left_arrow","t_right_arrow","c_left_leader","c_right_leader","c_left_arrow","c_right_arrow"],"does":[[139.813,"compression_caption (the \"Compression\" part) is emphasized."],[139.813,"tension_caption (the \"Tension\" part) is emphasized."]]},{"start":148.1915,"say":"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.","live":null,"does":[[148.1915,"compression_caption (the \"Compression\" part) is no longer emphasized."],[148.1915,"tension_caption (the \"Tension\" part) is no longer emphasized."],[159.95041666666665,"compression_caption is hidden from the screen — left the board."],[159.95041666666665,"compression_case is hidden from the screen — left the board."],[159.95041666666665,"c_bar is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_left_joint is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_right_joint is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_left_leader is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_right_leader is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_left_arrow is hidden from the screen — compression_case left the board."],[159.95041666666665,"c_right_arrow is hidden from the screen — compression_case left the board."],[159.95041666666665,"head_tc is hidden from the screen — left the board."],[159.95041666666665,"tension_caption is hidden from the screen — left the board."],[159.95041666666665,"tension_case is hidden from the screen — left the board."],[159.95041666666665,"t_bar is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_left_joint is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_right_joint is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_left_leader is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_right_leader is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_left_arrow is hidden from the screen — tension_case left the board."],[159.95041666666665,"t_right_arrow is hidden from the screen — tension_case left the board."]]}]},{"title":"Counting Before Solving","start":160.9920833333333,"end":291.4241666666666,"objects":{"bottom_left":"a Line [blue] drawn in truss (end=(4.0, 0.0))","bottom_right":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","head_count":"a Heading that says \"Eight Unknowns, Eight Equations\"","head_rule":"a Heading that says \"The Counting Rule\"","joint_a":"a Point [text] labelled \"A\" drawn in truss","joint_b":"a Point [text] labelled \"B\" drawn in truss (location=(4.0, 0.0))","joint_c":"a Point [text] labelled \"C\" drawn in truss (location=(8.0, 0.0))","joint_d":"a Point [text] labelled \"D\" drawn in truss (location=(4.0, 3.0))","lab_ab":"a Math [text] that says \"$F_(A B)$\" drawn in truss","lab_ad":"a Math [text] that says \"$F_(A D)$\" drawn in truss","lab_bc":"a Math [text] that says \"$F_(B C)$\" drawn in truss","lab_bd":"a Math [text] that says \"$F_(B D)$\" drawn in truss","lab_dc":"a Math [text] that says \"$F_(D C)$\" drawn in truss","left_diagonal":"a Line [blue] drawn in truss (end=(4.0, 3.0))","load":"a Vector [magenta] labelled \"12 thin upright(\"kN\")\" drawn in truss (start=(4.0, 4.9), end=(4.0, 3.3))","outcomes":"a Block [text] that says \"$m + r < 2 j$: too few bars. 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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.","live":["truss","head_count","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","pin_plate","pin_ground","roller_plate","roller_left","roller_right","roller_ground"],"does":[[172.70308333333332,"lab_ab is shown on the screen, written out."],[172.95308333333332,"lab_bc is shown on the screen, written out."],[173.20308333333332,"lab_ad is shown on the screen, written out."],[173.45308333333332,"lab_dc is shown on the screen, written out."],[173.70308333333332,"lab_bd is shown on the screen, written out."]]},{"start":182.6485833333333,"say":"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.","live":["truss","head_count","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","pin_plate","pin_ground","roller_plate","roller_left","roller_right","roller_ground","lab_ab","lab_bc","lab_ad","lab_dc","lab_bd"],"does":[[184.8660833333333,"pin_plate is hidden from the screen."],[184.8660833333333,"pin_ground is hidden from the screen."],[184.8660833333333,"roller_plate is hidden from the screen."],[184.8660833333333,"roller_left is hidden from the screen."],[184.8660833333333,"roller_right is hidden from the screen."],[184.8660833333333,"roller_ground is hidden from the screen."],[190.38108333333332,"react_ax is shown on the screen, written out."],[191.6350833333333,"react_ay is shown on the screen, written out."],[195.8960833333333,"react_cy is shown on the screen, written out."]]},{"start":198.93408333333332,"say":"Five bar forces and three reaction components. Eight unknown numbers in total, and that is the left hand side of the comparison.","live":["truss","head_count","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","lab_ab","lab_bc","lab_ad","lab_dc","lab_bd","react_ax","react_ay","react_cy"],"does":[[198.93408333333332,"truss moves to a new place on the board."],[198.93408333333332,"tally_table is shown on the screen, written out."],[199.63108333333332,"tally_table is shown on the screen, written out."],[200.8150833333333,"tally_table is shown on the screen, written out."],[202.6380833333333,"tally is shown on the screen, written out."]]},{"start":207.62608333333333,"say":"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.","live":null,"does":[[212.0730833333333,"joint_b is indicated — a transient flash."],[215.47508333333332,"react_ay is indicated — a transient flash."]]},{"start":222.1930833333333,"say":"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.","live":null,"does":[[231.84008333333333,"tally_table is shown on the screen, written out."],[234.4760833333333,"tally is shown on the screen, written out."]]},{"start":236.3300833333333,"say":"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.","live":null,"does":[[239.73208333333332,"tally is shown on the screen, written out."],[242.1110833333333,"A box is drawn around tally."],[249.5650833333333,"head_count is hidden from the screen — left the board."],[249.5650833333333,"tally is hidden from the screen — left the board."],[249.5650833333333,"tally_table is hidden from the screen — left the board."]]},{"start":250.7650833333333,"say":"It is worth knowing what the other two answers would have meant, because you will meet both.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","lab_ab","lab_bc","lab_ad","lab_dc","lab_bd","react_ax","react_ay","react_cy"],"does":[[250.7650833333333,"head_rule is shown on the screen, written out."],[252.72708333333333,"outcomes is shown on the screen, written out."]]},{"start":256.2295833333333,"say":"If the unknowns fall short of the equations, there are too few bars to hold the shape. 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But it has to start somewhere, and joints are not where it starts.","live":null,"does":[[282.0540833333333,"outcomes (the \"statically determinate\" part) is emphasized."],[282.0540833333333,"outcomes (the \"statically indeterminate\" part) is no longer emphasized."],[288.3810833333333,"outcomes (the \"statically determinate\" part) is no longer emphasized."],[290.38249999999994,"head_rule is hidden from the screen — left the board."],[290.38249999999994,"outcomes is hidden from the screen — left the board."],[290.38249999999994,"truss is hidden from the screen — left the board."],[290.38249999999994,"bottom_left is hidden from the screen — truss left the board."],[290.38249999999994,"bottom_right is hidden from the screen — truss left the board."],[290.38249999999994,"left_diagonal is hidden from the screen — truss left the board."],[290.38249999999994,"right_diagonal is hidden from the screen — truss left the board."],[290.38249999999994,"vertical is hidden from the screen — truss left the board."],[290.38249999999994,"joint_a is hidden from the screen — truss left the board."],[290.38249999999994,"joint_b is hidden from the screen — truss left the board."],[290.38249999999994,"joint_c is hidden from the screen — truss left the board."],[290.38249999999994,"joint_d is hidden from the screen — truss left the board."],[290.38249999999994,"load is hidden from the screen — truss left the board."],[290.38249999999994,"lab_ab is hidden from the screen — truss left the board."],[290.38249999999994,"lab_bc is hidden from the screen — truss left the board."],[290.38249999999994,"lab_ad is hidden from the screen — truss left the board."],[290.38249999999994,"lab_dc is hidden from the screen — truss left the board."],[290.38249999999994,"lab_bd is hidden from the screen — truss left the board."],[290.38249999999994,"react_ax is hidden from the screen — truss left the board."],[290.38249999999994,"react_ay is hidden from the screen — truss left the board."],[290.38249999999994,"react_cy is hidden from the screen — truss left the board."]]}]},{"title":"The Support Reactions","start":291.4241666666666,"end":442.51745833333325,"objects":{"arm_cy":"a Line [gray] labelled \"8 thin upright(\"m\")\" drawn in truss (start=(0.0, -2.48), end=(8.0, -2.48), dashed=True)","arm_load":"a Line [gray] labelled \"4 thin upright(\"m\")\" drawn in truss (start=(0.0, -2.05), end=(4.0, -2.05), dashed=True)","bottom_left":"a Line [blue] drawn in truss (end=(4.0, 0.0))","bottom_right":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","head_react":"a Heading that says \"The Whole Truss as One Body\"","head_ready":"a Heading that says \"The Doorway Into the Truss\"","joint_a":"a Point [text] labelled \"A\" drawn in truss","joint_b":"a Point [text] labelled \"B\" drawn in truss (location=(4.0, 0.0))","joint_c":"a Point [text] labelled \"C\" drawn in truss (location=(8.0, 0.0))","joint_d":"a Point [text] labelled \"D\" drawn in truss (location=(4.0, 3.0))","left_diagonal":"a Line [blue] drawn in truss (end=(4.0, 3.0))","load":"a Vector [magenta] labelled \"12 thin upright(\"kN\")\" drawn in truss (start=(4.0, 4.9), end=(4.0, 3.3))","react_ax":"a Vector [yellow] labelled \"A_x\" drawn in truss (start=(-1.55, 0.0), end=(-0.18, 0.0))","react_ay":"a Vector [yellow] labelled \"A_y\" drawn in truss (start=(0.0, -1.75), end=(0.0, -0.18))","react_cy":"a Vector [yellow] labelled \"C_y\" drawn in truss (start=(8.0, -1.75), end=(8.0, -0.18))","ready_note":"a Text [text] that says \"Every joint we visit from here has at most two unknown bar forces left, and exactly two equations to spend on them.\"","right_diagonal":"a Line [blue] drawn in truss (start=(4.0, 3.0), end=(8.0, 0.0))","truss":"a Figure (x_range=(-1.8, 9.8), y_range=(-2.6, 5.4), aspect=(11.6, 8.0))","vertical":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(4.0, 3.0))","work":"a Derivation [text] that says \"$sum M_A &= 0 \\ 8 C_y - 12 (4) &= 0 \\ C_y &= 6 thin upright(\"kN\") \\ sum F_y &= 0 \\ A_y + 6 - 12 &= 0 \\ A_y &= 6 thin upright(\"kN\") \\ sum F_x &= 0 \\ A_x &= 0$\""},"beats":[{"start":291.4241666666666,"say":"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.","live":[],"does":[[291.4241666666666,"head_react is shown on the screen, written out."],[291.4241666666666,"truss is shown on the screen, written out."],[292.05116666666663,"bottom_left is shown on the screen, written out."],[292.2511666666666,"bottom_right is shown on the screen, written out."],[292.4511666666666,"left_diagonal is shown on the screen, written out."],[292.6511666666666,"right_diagonal is shown on the screen, written out."],[292.85116666666664,"vertical is shown on the screen, written out."],[297.7171666666666,"joint_a is shown on the screen, written out."],[297.8671666666666,"joint_b is shown on the screen, written out."],[298.01716666666664,"joint_c is shown on the screen, written out."],[298.1671666666666,"joint_d is shown on the screen, written out."]]},{"start":305.7246666666666,"say":"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.","live":["truss","head_react","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d"],"does":[[310.4491666666666,"load is shown on the screen, written out."],[311.3551666666666,"react_ay is shown on the screen, written out."],[311.6551666666666,"react_ax is shown on the screen, written out."],[311.9551666666666,"react_cy is shown on the screen, written out."]]},{"start":317.1911666666666,"say":"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.","live":["truss","head_react","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_ax","react_cy"],"does":[]},{"start":334.23116666666664,"say":"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.","live":null,"does":[[334.8811666666666,"truss moves to a new place on the board."],[334.8811666666666,"work is shown on the screen, written out."],[343.4031666666666,"react_ax is indicated — a transient flash."],[343.4031666666666,"react_ay is indicated — a transient flash."]]},{"start":350.7831666666666,"say":"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.","live":null,"does":[[355.2761666666666,"arm_load is shown on the screen, written out."],[361.8121666666666,"arm_cy is shown on the screen, written out."]]},{"start":365.2681666666666,"say":"Set the sum to zero. Eight times C y, minus twelve times four, equals nothing.","live":["truss","head_react","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_ax","react_cy","arm_load","arm_cy"],"does":[[365.57016666666664,"work is shown on the screen, written out."]]},{"start":372.7876666666666,"say":"Twelve times four is forty eight, so C y is six kilonewtons, pushing upward.","live":null,"does":[[376.4571666666666,"work is shown on the screen, written out."],[378.14016666666663,"react_cy: one name gives way to another over the same drawing (label_becomes)."],[378.14016666666663,"arm_load is hidden from the screen."],[378.14016666666663,"arm_cy is hidden from the screen."]]},{"start":379.5876666666666,"say":"Now vertical forces. Six kilonewtons up at C, twelve down at the peak, and A y, whatever it turns out to be.","live":["truss","head_react","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_ax","react_cy"],"does":[[380.2381666666666,"work is shown on the screen, written out."],[386.3561666666666,"work is shown on the screen, written out."]]},{"start":388.6746666666666,"say":"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.","live":null,"does":[[390.1371666666666,"work is shown on the screen, written out."],[391.1361666666666,"react_ay: one name gives way to another over the same drawing (label_becomes)."]]},{"start":400.1766666666666,"say":"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.","live":null,"does":[[400.3971666666666,"work is shown on the screen, written out."]]},{"start":413.0246666666666,"say":"Which makes it zero. So we can rub that arrow out and never think about it again.","live":null,"does":[[414.04616666666664,"work is shown on the screen, written out."],[415.8341666666666,"react_ax is hidden from the screen."],[418.6671666666666,"truss moves to a new place on the board."],[418.6671666666666,"head_react is hidden from the screen — left the board."],[418.6671666666666,"work is hidden from the screen — left the board."]]},{"start":419.8671666666666,"say":"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.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy"],"does":[[419.8671666666666,"head_ready is shown on the screen, written out."],[420.8771666666666,"react_ay is indicated — a transient flash."],[420.8771666666666,"react_cy is indicated — a transient flash."]]},{"start":429.8936666666666,"say":"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.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","head_ready"],"does":[[430.89216666666664,"ready_note is shown on the screen, written out."],[441.4757916666666,"head_ready is hidden from the screen — left the board."],[441.4757916666666,"ready_note is hidden from the screen — left the board."],[441.4757916666666,"truss is hidden from the screen — left the board."],[441.4757916666666,"bottom_left is hidden from the screen — truss left the board."],[441.4757916666666,"bottom_right is hidden from the screen — truss left the board."],[441.4757916666666,"left_diagonal is hidden from the screen — truss left the board."],[441.4757916666666,"right_diagonal is hidden from the screen — truss left the board."],[441.4757916666666,"vertical is hidden from the screen — truss left the board."],[441.4757916666666,"joint_a is hidden from the screen — truss left the board."],[441.4757916666666,"joint_b is hidden from the screen — truss left the board."],[441.4757916666666,"joint_c is hidden from the screen — truss left the board."],[441.4757916666666,"joint_d is hidden from the screen — truss left the board."],[441.4757916666666,"load is hidden from the screen — truss left the board."],[441.4757916666666,"react_ay is hidden from the screen — truss left the board."],[441.4757916666666,"react_cy is hidden from the screen — truss left the board."]]}]},{"title":"Joint A, and What the Sign Means","start":442.51745833333325,"end":662.9070833333332,"objects":{"ab_tension":"a Line [green] drawn in truss (end=(4.0, 0.0))","ad_compression":"a Line [red] drawn in truss (end=(4.0, 3.0))","back_note":"a Text [text] that says \"A diagonal in compression shoves the base outward, and the bottom chord is stretched holding it in.\"","bottom_left":"a Line [blue] drawn in truss (end=(4.0, 0.0))","bottom_right":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","convention":"a Panel that says \"Draw every unknown bar force pulling away from the joint. A positive answer means tension. A negative answer means the bar is pushing, which is compression.\"","force_ab":"a Vector [green] labelled \"F_(A B)\" drawn in joint (start=(0.2, 0.0), end=(2.4, 0.0))","force_ad":"a Vector [green] labelled \"F_(A D)\" drawn in joint (start=(0.16, 0.12), end=(1.92, 1.44))","head_back":"a Heading that says \"Two Bars Down, Three to Go\"","head_convention":"a Heading that says \"Joint A, Cut Out\"","head_second":"a Heading that says \"Joint A: Then Horizontal\"","head_start":"a Heading that says \"Joint A: Vertical First\"","joint":"a Figure (x_range=(-2.6, 3.6), y_range=(-2.2, 2.8), aspect=(6.2, 5.0))","joint_a":"a Point [text] labelled \"A\" drawn in truss","joint_b":"a Point [text] labelled \"B\" drawn in truss (location=(4.0, 0.0))","joint_c":"a Point [text] labelled \"C\" drawn in truss (location=(8.0, 0.0))","joint_d":"a Point [text] labelled \"D\" drawn in truss (location=(4.0, 3.0))","lab_ab":"a Math [text] that says \"$F_(A B)$\" drawn in truss","lab_ad":"a Math [text] that says \"$F_(A D)$\" drawn in truss","lab_bc":"a Math [text] that says \"$F_(B C)$\" drawn in truss","lab_bd":"a Math [text] that says \"$F_(B D)$\" drawn in truss","lab_dc":"a Math [text] that says \"$F_(D C)$\" drawn in truss","left_diagonal":"a Line [blue] drawn in truss (end=(4.0, 3.0))","load":"a Vector [magenta] labelled \"12 thin upright(\"kN\")\" drawn in truss (start=(4.0, 4.9), end=(4.0, 3.3))","pivot":"a Point [text] labelled \"A\" drawn in joint","push_ad":"a Vector [red] labelled \"10 thin upright(\"kN\")\" drawn in joint (start=(1.92, 1.44), end=(0.16, 0.12))","react_ay":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in truss (start=(0.0, -1.75), end=(0.0, -0.18))","react_cy":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in truss (start=(8.0, -1.75), end=(8.0, -0.18))","reaction":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in joint (start=(0.0, -1.75), end=(0.0, -0.2))","result_ab":"a Math [text] that says \"$F_(A B) = 8 thin upright(\"kN\") thin (upright(\"T\"))$\"","result_ad":"a Math [text] that says \"$F_(A D) = 10 thin upright(\"kN\") thin (upright(\"C\"))$\"","right_diagonal":"a Line [blue] drawn in truss (start=(4.0, 3.0), end=(8.0, 0.0))","rise":"a Line [gray] drawn in joint (start=(1.92, 1.44), end=(1.92, 0.0), dashed=True)","square":"an Angle [gray] drawn in joint (vertex=(1.92, 0.0), sides=((1.92, 1.44), (0.0, 0.0)), right_angle=True)","theta":"an Angle [gray] labelled \"theta\" drawn in joint (sides=((1.4, 0.0), (1.12, 0.84)), radius=0.6)","truss":"a Figure (x_range=(-1.8, 9.8), y_range=(-2.6, 5.4), aspect=(11.6, 8.0))","vertical":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(4.0, 3.0))","work_x":"a Derivation [text] that says \"$sum F_x &= 0 \\ F_(A B) + F_(A D) cos theta &= 0 \\ F_(A B) + frac(4, 5) (-10) &= 0 \\ F_(A B) &= +8 thin upright(\"kN\")$\"","work_y":"a Derivation [text] that says \"$sum F_y &= 0 \\ 6 + F_(A D) sin theta &= 0 \\ 6 + frac(3, 5) F_(A D) &= 0 \\ F_(A D) &= -10 thin upright(\"kN\")$\""},"beats":[{"start":442.51745833333325,"say":"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.","live":[],"does":[[442.51745833333325,"head_convention is shown on the screen, written out."],[442.51745833333325,"joint is shown on the screen, written out."],[450.35445833333324,"pivot is shown on the screen, written out."]]},{"start":454.49545833333326,"say":"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.","live":["joint","head_convention","pivot"],"does":[[458.88345833333324,"reaction is shown on the screen, written out."]]},{"start":464.57995833333325,"say":"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.","live":["joint","head_convention","pivot","reaction"],"does":[]},{"start":477.1029583333333,"say":"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.","live":null,"does":[[482.32745833333325,"force_ab is shown on the screen, written out."],[482.32745833333325,"force_ad is shown on the screen, written out."],[486.62345833333325,"joint moves to a new place on the board."],[486.62345833333325,"convention is shown on the screen, written out."]]},{"start":488.1174583333333,"say":"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.","live":["convention","joint","head_convention","pivot","reaction","force_ab","force_ad"],"does":[[496.88245833333326,"convention (the \"positive answer means tension\" part) is emphasized."],[503.5234583333332,"convention (the \"positive answer means tension\" part) is no longer emphasized."],[503.5234583333332,"convention (the \"the bar is pushing\" part) is emphasized."]]},{"start":504.85495833333323,"say":"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.","live":null,"does":[[504.85495833333323,"convention (the \"the bar is pushing\" part) is no longer emphasized."],[513.2374583333333,"theta is shown on the screen, written out."],[515.1294583333332,"rise is shown on the screen, written out."],[515.1294583333332,"square is shown on the screen, written out."],[518.5424583333332,"convention is hidden from the screen — left the board."],[518.5424583333332,"head_convention is hidden from the screen — left the board."]]},{"start":519.7424583333333,"say":"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.","live":["joint","pivot","reaction","force_ab","force_ad","theta","rise","square"],"does":[[519.7424583333333,"head_start is shown on the screen, written out."],[520.4854583333332,"work_y is shown on the screen, written out."],[525.3274583333332,"force_ab is indicated — a transient flash."]]},{"start":534.0189583333332,"say":"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.","live":["joint","pivot","reaction","force_ab","force_ad","theta","rise","square","head_start"],"does":[[535.5164583333333,"work_y is shown on the screen, written out."],[545.7104583333332,"work_y is shown on the screen, written out."]]},{"start":547.3089583333333,"say":"Three fifths of F A D has to cancel six, so F A D is minus ten kilonewtons.","live":null,"does":[[552.1734583333332,"work_y is shown on the screen, written out."]]},{"start":554.6139583333332,"say":"And there is the negative. It is not a bookkeeping nuisance, it is the answer telling us something. The arrow we drew is backwards.","live":null,"does":[[555.5024583333333,"work_y (the \"-10\" part) is emphasized."]]},{"start":563.7474583333333,"say":"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.","live":null,"does":[[563.7474583333333,"work_y (the \"-10\" part) is no longer emphasized."],[567.8984583333332,"force_ad is hidden from the screen."],[567.8984583333332,"rise is hidden from the screen."],[567.8984583333332,"square is hidden from the screen."],[569.7904583333333,"push_ad is shown on the screen, written out."],[573.4009583333333,"head_start is hidden from the screen — left the board."],[573.4009583333333,"work_y is hidden from the screen — left the board."]]},{"start":574.6009583333332,"say":"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.","live":["joint","pivot","reaction","force_ab","theta","push_ad"],"does":[[574.6009583333332,"head_second is shown on the screen, written out."],[576.0284583333332,"result_ad is shown on the screen, written out."],[581.5554583333333,"work_x is shown on the screen, written out."]]},{"start":584.8719583333332,"say":"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.","live":["joint","pivot","reaction","force_ab","theta","push_ad","result_ad","head_second"],"does":[[588.5984583333333,"work_x is shown on the screen, written out."]]},{"start":596.9999583333332,"say":"Four fifths of minus ten is minus eight, pointing to the left. So F A B must be plus eight kilonewtons to balance it.","live":null,"does":[[599.5084583333332,"work_x is shown on the screen, written out."],[605.2544583333332,"work_x is shown on the screen, written out."]]},{"start":606.8184583333332,"say":"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.","live":null,"does":[[607.1204583333332,"result_ab is shown on the screen, written out."],[612.4264583333332,"force_ab is indicated — a transient flash."]]},{"start":616.9039583333332,"say":"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.","live":["joint","pivot","reaction","force_ab","theta","push_ad","result_ad","result_ab","head_second"],"does":[[622.8714583333333,"push_ad is indicated — a transient flash."],[629.5244583333332,"force_ab is indicated — a transient flash."],[630.5579583333332,"head_second is hidden from the screen — left the board."],[630.5579583333332,"joint is hidden from the screen — left the board."],[630.5579583333332,"pivot is hidden from the screen — joint left the board."],[630.5579583333332,"reaction is hidden from the screen — joint left the board."],[630.5579583333332,"force_ab is hidden from the screen — joint left the board."],[630.5579583333332,"theta is hidden from the screen — joint left the board."],[630.5579583333332,"push_ad is hidden from the screen — joint left the board."],[630.5579583333332,"result_ab is hidden from the screen — left the board."],[630.5579583333332,"result_ad is hidden from the screen — left the board."],[630.5579583333332,"work_x is hidden from the screen — left the board."]]},{"start":631.7579583333331,"say":"Step back to the whole truss, with the reactions we found and the three bar forces we have not touched yet.","live":[],"does":[[631.7579583333331,"head_back is shown on the screen, written out."],[631.7579583333331,"truss is shown on the screen, written out."],[633.2204583333332,"bottom_left is shown on the screen, written out."],[633.4004583333332,"bottom_right is shown on the screen, written out."],[633.5804583333332,"left_diagonal is shown on the screen, written out."],[633.7604583333332,"right_diagonal is shown on the screen, written out."],[633.9404583333333,"vertical is shown on the screen, written out."],[634.3354583333332,"joint_a is shown on the screen, written out."],[634.3354583333332,"load is shown on the screen, written out."],[634.4754583333332,"joint_b is shown on the screen, written out."],[634.6154583333332,"joint_c is shown on the screen, written out."],[634.7554583333332,"joint_d is shown on the screen, written out."],[635.1474583333332,"react_ay is shown on the screen, written out."],[635.1474583333332,"react_cy is shown on the screen, written out."],[637.6444583333332,"lab_bc is shown on the screen, written out."],[637.8444583333333,"lab_bd is shown on the screen, written out."],[638.0444583333332,"lab_dc is shown on the screen, written out."]]},{"start":639.4284583333332,"say":"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.","live":["truss","head_back","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","lab_bc","lab_bd","lab_dc"],"does":[[644.9434583333332,"ad_compression is shown on the screen, written out."],[644.9434583333332,"lab_ad is shown on the screen, written out."],[645.3434583333333,"lab_ad becomes \"$10 thin upright(\"kN\") thin (upright(\"C\"))$\"."],[649.7384583333333,"ab_tension is shown on the screen, written out."],[649.7384583333333,"lab_ab is shown on the screen, written out."],[650.1384583333332,"lab_ab becomes \"$8 thin upright(\"kN\") thin (upright(\"T\"))$\"."]]},{"start":651.1854583333331,"say":"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.","live":["truss","head_back","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","lab_bc","lab_bd","lab_dc","ad_compression","lab_ad","ab_tension","lab_ab"],"does":[[651.1854583333331,"back_note is shown on the screen, written out."],[661.8654166666665,"back_note is hidden from the screen — left the board."],[661.8654166666665,"head_back is hidden from the screen — left the board."],[661.8654166666665,"truss is hidden from the screen — left the board."],[661.8654166666665,"bottom_left is hidden from the screen — truss left the board."],[661.8654166666665,"bottom_right is hidden from the screen — truss left the board."],[661.8654166666665,"left_diagonal is hidden from the screen — truss left the board."],[661.8654166666665,"right_diagonal is hidden from the screen — truss left the board."],[661.8654166666665,"vertical is hidden from the screen — truss left the board."],[661.8654166666665,"joint_a is hidden from the screen — truss left the board."],[661.8654166666665,"joint_b is hidden from the screen — truss left the board."],[661.8654166666665,"joint_c is hidden from the screen — truss left the board."],[661.8654166666665,"joint_d is hidden from the screen — truss left the board."],[661.8654166666665,"load is hidden from the screen — truss left the board."],[661.8654166666665,"react_ay is hidden from the screen — truss left the board."],[661.8654166666665,"react_cy is hidden from the screen — truss left the board."],[661.8654166666665,"lab_bc is hidden from the screen — truss left the board."],[661.8654166666665,"lab_bd is hidden from the screen — truss left the board."],[661.8654166666665,"lab_dc is hidden from the screen — truss left the board."],[661.8654166666665,"ad_compression is hidden from the screen — truss left the board."],[661.8654166666665,"lab_ad is hidden from the screen — truss left the board."],[661.8654166666665,"ab_tension is hidden from the screen — truss left the board."],[661.8654166666665,"lab_ab is hidden from the screen — truss left the board."]]}]},{"title":"The Rest of the Truss","start":662.9070833333332,"end":935.6967083333332,"objects":{"ab_tension":"a Line [green] drawn in truss (end=(4.0, 0.0))","ad_compression":"a Line [red] drawn in truss (end=(4.0, 3.0))","b_left":"a Vector [green] labelled \"F_(A B)\" drawn in joint_b_fig (start=(-0.2, 0.0), end=(-2.2, 0.0))","b_note":"a Math [gray] that says \"$upright(\"no load applied here\")$\" drawn in joint_b_fig","b_pivot":"a Point [text] labelled \"B\" drawn in joint_b_fig","b_right":"a Vector [green] labelled \"F_(B C)\" drawn in joint_b_fig (start=(0.2, 0.0), end=(2.2, 0.0))","b_up":"a Vector [green] labelled \"F_(B D)\" drawn in joint_b_fig (start=(0.0, 0.2), end=(0.0, 2.3))","bc_tension":"a Line [green] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","bottom_left":"a Line [blue] drawn in truss (end=(4.0, 0.0))","bottom_right":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(8.0, 0.0))","c_chord":"a Vector [green] labelled \"F_(B C)\" drawn in joint_c_fig (start=(-0.2, 0.0), end=(-2.2, 0.0))","c_diagonal":"a Vector [green] labelled \"F_(C D)\" drawn in joint_c_fig (start=(-0.16, 0.12), end=(-1.92, 1.44))","c_pivot":"a Point [text] labelled \"C\" drawn in joint_c_fig","c_push":"a Vector [red] labelled \"10 thin upright(\"kN\")\" drawn in joint_c_fig (start=(-1.92, 1.44), end=(-0.16, 0.12))","c_reaction":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in joint_c_fig (start=(0.0, -1.7), end=(0.0, -0.2))","c_theta":"an Angle [gray] labelled \"theta\" drawn in joint_c_fig (sides=((-1.4, 0.0), (-1.12, 0.84)), radius=0.6)","dc_compression":"a Line [red] drawn in truss (start=(4.0, 3.0), end=(8.0, 0.0))","head_done":"a Heading that says \"The Finished Truss\"","head_jb":"a Heading that says \"Joint B: Spotted by Inspection\"","head_jc":"a Heading that says \"Joint C: The Mirror Image\"","head_jd":"a Heading that says \"Joint D: The Free Check\"","joint_a":"a Point [text] labelled \"A\" drawn in truss","joint_b":"a Point [text] labelled \"B\" drawn in truss (location=(4.0, 0.0))","joint_b_fig":"a Figure (x_range=(-2.8, 2.8), y_range=(-1.8, 3.2), aspect=(5.6, 5.0))","joint_c":"a Point [text] labelled \"C\" drawn in truss (location=(8.0, 0.0))","joint_c_fig":"a Figure (x_range=(-3.6, 2.4), y_range=(-2.0, 3.0), aspect=(6.0, 5.0))","joint_d":"a Point [text] labelled \"D\" drawn in truss (location=(4.0, 3.0))","lab_ab":"a Math [text] that says \"$8 thin upright(\"kN\") thin (upright(\"T\"))$\" drawn in truss","lab_ad":"a Math [text] that says \"$10 thin upright(\"kN\") thin (upright(\"C\"))$\" drawn in truss","lab_bc":"a Math [text] that says \"$8 thin upright(\"kN\") thin (upright(\"T\"))$\" drawn in truss","lab_bd":"a Math [text] that says \"$0$\" drawn in truss","lab_dc":"a Math [text] that says \"$10 thin upright(\"kN\") thin (upright(\"C\"))$\" drawn in truss","left_diagonal":"a Line [blue] drawn in truss (end=(4.0, 3.0))","load":"a Vector [magenta] labelled \"12 thin upright(\"kN\")\" drawn in truss (start=(4.0, 4.9), end=(4.0, 3.3))","react_ay":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in truss (start=(0.0, -1.75), end=(0.0, -0.18))","react_cy":"a Vector [yellow] labelled \"6 thin upright(\"kN\")\" drawn in truss (start=(8.0, -1.75), end=(8.0, -0.18))","results":"a Table [text] that says \"Bar Force State $A B$ 8 kN T $B C$ 8 kN T $A D$ 10 kN C $D C$ 10 kN C $B D$ 0 zero\" (rows=(('Bar', 'Force', 'State'), ('$A B$', '8 kN', 'T'), ('$B C$', '…, header=True)","right_diagonal":"a Line [blue] drawn in truss (start=(4.0, 3.0), end=(8.0, 0.0))","truss":"a Figure (x_range=(-1.8, 9.8), y_range=(-2.6, 5.4), aspect=(11.6, 8.0))","vertical":"a Line [blue] drawn in truss (start=(4.0, 0.0), end=(4.0, 3.0))","work_b":"a Derivation [text] that says \"$sum F_y &= 0 \\ F_(B D) &= 0 \\ sum F_x &= 0 \\ F_(B C) - F_(A B) &= 0 \\ F_(B C) &= 8 thin upright(\"kN\")$\"","work_c":"a Derivation [text] that says \"$sum F_y &= 0 \\ 6 + frac(3, 5) F_(C D) &= 0 \\ F_(C D) &= -10 thin upright(\"kN\") \\ sum F_x &= 0 \\ - F_(B C) - frac(4, 5) F_(C D) &= 0 \\ -8 + 8 &= 0$\"","work_d":"a Derivation [text] that says \"$sum F_x &= 0 \\ frac(4, 5) (10) - frac(4, 5) (10) &= 0 \\ sum F_y &= 0 \\ frac(3, 5) (10) + frac(3, 5) (10) - 12 &= 0$\"","zero_note":"a Panel that says \"Three bars meet at an unloaded joint and two of them are collinear. The third bar carries no force.\""},"beats":[{"start":662.9070833333332,"say":"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.","live":[],"does":[[662.9070833333332,"head_jb is shown on the screen, written out."],[662.9070833333332,"joint_b_fig is shown on the screen, written out."],[664.6480833333331,"b_pivot is shown on the screen, written out."],[667.8640833333332,"b_left is shown on the screen, written out."],[669.6990833333332,"b_right is shown on the screen, written out."],[671.8350833333332,"b_up is shown on the screen, written out."],[673.3330833333332,"b_note is shown on the screen, written out."]]},{"start":676.2080833333332,"say":"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.","live":["joint_b_fig","head_jb","b_pivot","b_left","b_right","b_up","b_note"],"does":[[682.2220833333331,"b_left is emphasized."],[682.2220833333331,"b_right is emphasized."]]},{"start":688.3255833333332,"say":"So the vertical bar is the only thing at this joint with any vertical component whatsoever. There is nothing for it to balance against.","live":null,"does":[[688.3255833333332,"b_left is no longer emphasized."],[688.3255833333332,"b_right is no longer emphasized."],[689.9390833333332,"b_up is emphasized."]]},{"start":697.0640833333332,"say":"Which means vertical equilibrium at B gives F B D equals zero, immediately, with no arithmetic at all. Away it goes.","live":null,"does":[[697.0640833333332,"b_up is no longer emphasized."],[698.3060833333332,"joint_b_fig moves to a new place on the board."],[698.3060833333332,"work_b is shown on the screen, written out."],[701.5570833333331,"work_b is shown on the screen, written out."],[705.0750833333332,"b_up is hidden from the screen."]]},{"start":706.7775833333332,"say":"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.","live":["joint_b_fig","head_jb","b_pivot","b_left","b_right","b_note"],"does":[[709.0530833333331,"zero_note is shown on the screen, written out."]]},{"start":718.3495833333332,"say":"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.","live":["zero_note","joint_b_fig","head_jb","b_pivot","b_left","b_right","b_note"],"does":[]},{"start":731.4300833333332,"say":"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.","live":null,"does":[[732.1730833333331,"work_b is shown on the screen, written out."],[737.7580833333332,"work_b is shown on the screen, written out."],[740.4630833333332,"work_b is shown on the screen, written out."],[744.0615833333331,"head_jb is hidden from the screen — left the board."],[744.0615833333331,"joint_b_fig is hidden from the screen — left the board."],[744.0615833333331,"b_pivot is hidden from the screen — joint_b_fig left the board."],[744.0615833333331,"b_left is hidden from the screen — joint_b_fig left the board."],[744.0615833333331,"b_right is hidden from the screen — joint_b_fig left the board."],[744.0615833333331,"b_note is hidden from the screen — joint_b_fig left the board."],[744.0615833333331,"work_b is hidden from the screen — left the board."],[744.0615833333331,"zero_note is hidden from the screen — left the board."]]},{"start":745.2615833333332,"say":"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.","live":[],"does":[[745.2615833333332,"head_jc is shown on the screen, written out."],[745.2615833333332,"joint_c_fig is shown on the screen, written out."],[749.1280833333332,"c_pivot is shown on the screen, written out."],[750.0910833333331,"c_reaction is shown on the screen, written out."],[753.3310833333331,"c_chord is shown on the screen, written out."],[755.8380833333332,"c_diagonal is shown on the screen, written out."],[757.7310833333331,"c_theta is shown on the screen, written out."]]},{"start":759.9335833333332,"say":"Same convention, drawn pulling away from the joint. Same geometry too, because this diagonal is also a three four five, just mirrored.","live":["joint_c_fig","head_jc","c_pivot","c_reaction","c_chord","c_diagonal","c_theta"],"does":[]},{"start":769.2755833333332,"say":"Vertical equilibrium. Six upward, plus three fifths of F C D, equals zero. So F C D is minus ten kilonewtons.","live":null,"does":[[769.5310833333332,"joint_c_fig moves to a new place on the board."],[769.5310833333332,"work_c is shown on the screen, written out."],[771.5860833333331,"work_c is shown on the screen, written out."],[777.6350833333331,"work_c is shown on the screen, written out."]]},{"start":780.3250833333332,"say":"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.","live":null,"does":[[783.1690833333332,"c_diagonal is hidden from the screen."],[785.4330833333331,"c_push is shown on the screen, written out."]]},{"start":791.2350833333331,"say":"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.","live":["joint_c_fig","head_jc","c_pivot","c_reaction","c_chord","c_theta","c_push"],"does":[[797.1550833333331,"work_c is shown on the screen, written out."]]},{"start":804.5240833333331,"say":"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.","live":null,"does":[[808.2390833333332,"work_c is shown on the screen, written out."],[813.2430833333332,"work_c is shown on the screen, written out."],[816.9935833333332,"head_jc is hidden from the screen — left the board."],[816.9935833333332,"joint_c_fig is hidden from the screen — left the board."],[816.9935833333332,"c_pivot is hidden from the screen — joint_c_fig left the board."],[816.9935833333332,"c_reaction is hidden from the screen — joint_c_fig left the board."],[816.9935833333332,"c_chord is hidden from the screen — joint_c_fig left the board."],[816.9935833333332,"c_theta is hidden from the screen — joint_c_fig left the board."],[816.9935833333332,"c_push is hidden from the screen — joint_c_fig left the board."],[816.9935833333332,"work_c is hidden from the screen — left the board."]]},{"start":818.1935833333332,"say":"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.","live":[],"does":[[818.1935833333332,"head_jd is shown on the screen, written out."],[818.1935833333332,"truss is shown on the screen, written out."],[819.1920833333331,"bottom_left is shown on the screen, written out."],[819.3520833333332,"bottom_right is shown on the screen, written out."],[819.5120833333332,"left_diagonal is shown on the screen, written out."],[819.6720833333331,"right_diagonal is shown on the screen, written out."],[819.7720833333332,"joint_a is shown on the screen, written out."],[819.7720833333332,"load is shown on the screen, written out."],[819.7720833333332,"react_ay is shown on the screen, written out."],[819.7720833333332,"react_cy is shown on the screen, written out."],[819.8320833333331,"vertical is shown on the screen, written out."],[819.8920833333332,"joint_b is shown on the screen, written out."],[820.0120833333332,"joint_c is shown on the screen, written out."],[820.1320833333332,"joint_d is shown on the screen, written out."],[821.8850833333331,"ab_tension is shown on the screen, written out."],[821.8850833333331,"bc_tension is shown on the screen, written out."],[822.8610833333332,"lab_ab is shown on the screen, written out."],[822.8610833333332,"lab_bc is shown on the screen, written out."],[826.2740833333332,"ad_compression is shown on the screen, written out."],[826.2740833333332,"dc_compression is shown on the screen, written out."],[827.0410833333332,"lab_ad is shown on the screen, written out."],[827.0410833333332,"lab_dc is shown on the screen, written out."],[828.9210833333332,"lab_bd is shown on the screen, written out."]]},{"start":831.8205833333332,"say":"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.","live":["truss","head_jd","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","ab_tension","bc_tension","lab_ab","lab_bc","ad_compression","dc_compression","lab_ad","lab_dc","lab_bd"],"does":[[839.4360833333332,"joint_d is indicated — a transient flash."]]},{"start":843.4725833333332,"say":"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.","live":null,"does":[[850.7170833333332,"truss moves to a new place on the board."],[850.7170833333332,"work_d is shown on the screen, written out."],[858.5770833333331,"work_d is shown on the screen, written out."]]},{"start":860.2680833333332,"say":"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.","live":null,"does":[[860.7440833333332,"work_d is shown on the screen, written out."],[866.7930833333331,"work_d is shown on the screen, written out."],[870.3450833333332,"load is indicated — a transient flash."],[875.2220833333331,"head_jd is hidden from the screen — left the board."],[875.2220833333331,"work_d is hidden from the screen — left the board."]]},{"start":876.4220833333331,"say":"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.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","ab_tension","bc_tension","lab_ab","lab_bc","ad_compression","dc_compression","lab_ad","lab_dc","lab_bd"],"does":[[876.4220833333331,"head_done is shown on the screen, written out."],[878.8370833333331,"results is shown on the screen, written out."],[880.3810833333332,"results is shown on the screen, written out."],[882.5170833333332,"results is shown on the screen, written out."],[884.3280833333331,"results is shown on the screen, written out."],[884.8510833333331,"results is shown on the screen, written out."],[886.5220833333332,"results is shown on the screen, written out."]]},{"start":888.8290833333332,"say":"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.","live":["truss","bottom_left","bottom_right","left_diagonal","right_diagonal","vertical","joint_a","joint_b","joint_c","joint_d","load","react_ay","react_cy","ab_tension","bc_tension","lab_ab","lab_bc","ad_compression","dc_compression","lab_ad","lab_dc","lab_bd","head_done"],"does":[[889.7810833333332,"results (the \"column=3\" part) is emphasized."],[895.3660833333331,"results (the \"column=3\" part) is no longer emphasized."],[897.7110833333331,"ab_tension is indicated — a transient flash."],[897.7110833333331,"bc_tension is indicated — a transient flash."],[901.1360833333332,"ad_compression is indicated — a transient flash."],[901.1360833333332,"dc_compression is indicated — a transient flash."]]},{"start":903.6635833333331,"say":"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.","live":null,"does":[]},{"start":917.3355833333331,"say":"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.","live":null,"does":[[924.0110833333332,"results (the \"row=4\" part) is emphasized."],[927.3780833333332,"results (the \"row=4\" part) is no longer emphasized."],[927.3780833333332,"results (the \"row=5\" part) is emphasized."],[930.6870833333331,"results (the \"row=5\" part) is no longer emphasized."],[934.6550416666664,"head_done is hidden from the screen — left the board."],[934.6550416666664,"results is hidden from the screen — left the board."],[934.6550416666664,"truss is hidden from the screen — left the board."],[934.6550416666664,"bottom_left is hidden from the screen — truss left the board."],[934.6550416666664,"bottom_right is hidden from the screen — truss left the board."],[934.6550416666664,"left_diagonal is hidden from the screen — truss left the board."],[934.6550416666664,"right_diagonal is hidden from the screen — truss left the board."],[934.6550416666664,"vertical is hidden from the screen — truss left the board."],[934.6550416666664,"joint_a is hidden from the screen — truss left the board."],[934.6550416666664,"joint_b is hidden from the screen — truss left the board."],[934.6550416666664,"joint_c is hidden from the screen — truss left the board."],[934.6550416666664,"joint_d is hidden from the screen — truss left the board."],[934.6550416666664,"load is hidden from the screen — truss left the board."],[934.6550416666664,"react_ay is hidden from the screen — truss left the board."],[934.6550416666664,"react_cy is hidden from the screen — truss left the board."],[934.6550416666664,"ab_tension is hidden from the screen — truss left the board."],[934.6550416666664,"bc_tension is hidden from the screen — truss left the board."],[934.6550416666664,"lab_ab is hidden from the screen — truss left the board."],[934.6550416666664,"lab_bc is hidden from the screen — truss left the board."],[934.6550416666664,"ad_compression is hidden from the screen — truss left the board."],[934.6550416666664,"dc_compression is hidden from the screen — truss left the board."],[934.6550416666664,"lab_ad is hidden from the screen — truss left the board."],[934.6550416666664,"lab_dc is hidden from the screen — truss left the board."],[934.6550416666664,"lab_bd is hidden from the screen — truss left the board."]]}]}]},"durationSeconds":936,"chapters":[{"title":"A Truss, and What Its Bars Can Do","startSeconds":0,"narration":"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."},{"title":"Counting Before Solving","startSeconds":160.9920833333333,"narration":"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."},{"title":"The Support Reactions","startSeconds":291.4241666666666,"narration":"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."},{"title":"Joint A, and What the Sign Means","startSeconds":442.51745833333325,"narration":"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."},{"title":"The Rest of the Truss","startSeconds":662.9070833333332,"narration":"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."}]}}