{"version":1,"lectureId":"01M14TY71B7KJ7N3M9NE8BZATV","attempt":0,"publication":{"slug":"euler-buckling-columns","title":"Euler Buckling: Why Slender Columns Fail by Instability","subject":"engineering","summary":"A slender column gives way long before its material yields, and this lecture shows why. Starting from a pin ended column under axial load, we let the column bend, write moment equilibrium for the displaced shape, and solve the resulting differential equation to find that the straight configuration stops being the only equilibrium once the load reaches pi squared E I over L squared. The buckled mode shape is drawn, its amplitude is shown to be indeterminate, and the higher modes are located. Changing the end conditions then turns the effective length factor into a statement about where the bending moment vanishes, rather than a number looked up in a table. The lecture closes with the slenderness ratio and an honest account of where elastic buckling stops governing and plain yielding takes over.","metaDescription":"Why slender columns buckle: the critical load derived from the deflected shape, effective length from inflection points, and the slenderness limit.","transcript":"A steel strut one metre long will carry an enormous load. Make the same strut three metres long, from the same steel, with the same cross section, and it folds up under a small fraction of that. Nothing about the material changed. Here are two columns cut from one bar. The short one fails the way you expect. Push hard enough, the stress reaches the yield strength of the steel, and the material gives way. Now the long one. Push on it, and long before the stress anywhere in it comes near yield, it does this. It bows sideways. The material is still perfectly elastic. Unload it and it springs straight again. But as a structural member it has failed, because it will not take any more load than that. And the gap between the two is not small. Triple the length, and the load the column can take falls by a factor of nine. Nothing in the material noticed. This is buckling, and it is a failure of stability rather than a failure of strength. The question is what load it happens at, and why the length matters so much when the material does not. Here is the case we will solve. A straight column of length L, pinned at both ends, so each end is free to rotate but held against sideways movement. It carries an axial load P through the centroid. The material is linearly elastic with Young's modulus E, and the cross section has a second moment of area I about the axis it bends about. Those are the only four quantities the answer can depend on. The answer we are heading for is this. The critical load is pi squared E I over L squared, and the shape the column takes at that load is a single smooth half sine wave, zero at both pins and largest at midheight. I want both of those to fall out of the mechanics rather than be handed to you. And notice already what that formula does not contain. There is no yield stress in it anywhere. The load at which a slender column buckles has nothing to do with how strong the material is. Only with how stiff it is, and with how the length and the cross section combine. So let the column bend, write equilibrium for the bent shape, and see what the mathematics demands. Suppose the column is not straight. Not because something pushed it out of line, but suppose it simply is, by some small amount, and ask whether that bent shape can stand in equilibrium. The dashed grey line is where the axis used to be. Take a height x above the bottom pin, and call the sideways movement of the axis at that station v of x. Now cut the column there, and keep the piece below the cut. At the bottom, the axial load P acts along its original line of action, because that pin has not moved sideways at all. And there is no sideways reaction at the pin either, because there is no sideways load anywhere on the column. So the base hands the free body exactly one force, P, at zero offset from the original axis. That is the whole free body. One axial force at the base, and the internal shear and moment at the cut. Nothing else is acting on the piece. Take moments about the cut section. The load P is acting a distance v of x away from that section, so it applies a moment of P times v of x. The internal bending moment has to balance it exactly. So M of x plus P v of x is zero, and the bending moment at any section is minus the axial load, times how far that section has strayed from the line of action. Watch what that means as the cut moves. Near the ends the deflection is small, so the moment is small. At midheight the deflection is largest, and so is the bending moment. So the bending moment along the column has the same shape as the deflection itself, scaled by P. That is unusual. In ordinary beam bending you are handed the loading and you go and find the moment. Here the moment depends on the answer. And there is the loop that makes buckling what it is. Deflection produces bending moment. Bending moment produces curvature. Curvature produces more deflection. The load P is the gain around that loop. Now bring in the bending relation you already have. For small deflections, E I times the second derivative of the deflection equals the bending moment at that section. That is the small deflection form, and it is the honest limit of what we are doing here. As long as the bow stays small compared with the length, the curvature is well approximated by v double prime. Substitute the moment we just found. E I v double prime equals minus P v. Move everything to one side, and there is the governing equation. Divide through by E I, and give the group P over E I a name. Call it k squared. Then the equation reads v double prime plus k squared v equals zero, and k carries units of one over length. That is a second order linear equation with constant coefficients, and you already know every function that satisfies it. Solving it is what turns this into a statement about the load. v double prime plus k squared v equals zero. Every solution of that is a combination of a sine and a cosine of k x, so write the general one down with two constants in it. Now the end conditions. The bottom is pinned, so it cannot move sideways there. v of zero is zero. Put x equal to zero into the general solution: the sine term vanishes on its own, and what is left is B. So B is zero, and that leaves v equals A sine k x. The top is pinned too, so v of L is zero as well. Put x equal to L in, and you get A times sine of k L equals zero. Look hard at that line, because everything is in it. A product of two things is zero, so at least one of them has to be zero, and there are exactly two ways for that to happen. The first way is that A is zero. That gives v identically zero, the perfectly straight column. And notice it is available at every value of P. The straight configuration is always an equilibrium, and it never stops being one. The second way is that the sine itself is zero. Sine of k L vanishes when k L is a whole number of pi. And that is a condition on the load, because k squared is P over E I. Square both sides, substitute for k squared, and P L squared over E I equals n squared pi squared. So P equals n squared pi squared E I over L squared. These are not loads the column happens to reach. They are the only loads at which anything other than the straight shape is possible at all. Below the smallest of them, A has to be zero. The smallest is n equal to one. Pi squared E I over L squared. That is the Euler critical load, and it came out of demanding that a bent shape can stand in equilibrium, not out of any strength calculation. And the shape that goes with it is A sine of pi x over L. One half of a sine wave. Zero at both pins, largest at midheight, and no point of inflection anywhere in between. Now the part that catches people out. A is still undetermined. The mathematics has fixed the shape and said nothing whatever about how big it is. Here is the same mode at three different amplitudes. Every one of them satisfies equilibrium exactly, and every one of them does it at exactly the same load. That is the signature of the thing we are looking at. The higher values of n are genuine solutions too. n equal to two is a full sine wave, with a stationary point at midheight, and it needs four times the load. n equal to three needs nine times. In a bare column they never happen, because the column has already gone at n equal to one. But brace it sideways at midheight, and you have forbidden the first mode. The second one is what you get instead. Put all of that on one picture. Load runs up the vertical axis. The amplitude of the bow runs across the horizontal one, so the whole vertical axis is the straight column. Load it up from nothing. Below the critical load there is exactly one equilibrium at every value of P, and it sits on that axis at zero amplitude. At the critical load, a second branch appears. Every amplitude on that horizontal line is an equilibrium, at the one load, and the straight solution has stopped being unique. That splitting is called a bifurcation, and it is exactly what we mean when we say the column has buckled. Nothing broke. No stress reached any limit. The column simply ran out of reasons to stay straight. Now, that whole derivation used one particular pair of end conditions. Change them and the numbers change. What is worth seeing is that they change in a way you can read straight off the deflected shape. Everything in that derivation came from two facts about the ends. They could not move sideways, and they could not carry bending moment, because a pin cannot. Change either one and the answer changes. But look at what the pinned answer really is. The moment is P times the deflection, and the deflection is zero at both pins. So the half wave runs between two points where the bending moment is zero. So here is the statement that replaces the table. The critical load is pi squared E I divided by the square of the distance between adjacent points of zero bending moment. Call that distance the effective length. The reason is simple. Between two such points, the deflected curve is a half sine wave with no moment at either end. That is precisely the pin ended problem we already solved, sitting inside a longer column, and it does not care what happens beyond those two points. And zero bending moment means zero curvature, so those points are the inflection points of the deflected shape. You can find them by looking at where the curve changes the way it bends. Take a column built in at both ends, so neither end can rotate. Load it until it buckles, and it takes this shape: vertical where it meets each support, bulging one way in the middle, and curling back near each end. Find the inflection points. The curve bends one way near the middle and the other way close to each support. It changes over here, and again here, at a quarter of the length in from each end. So the distance between them is half the length of the column. That middle half is a half sine wave with zero moment at both ends, and it has no idea it is not a pinned column of length L over two. Put L over two in place of L. The critical load is four times what the pinned column would carry. The factor K is one half, and it is not a number from a table. It is where the curvature vanishes. Now the other extreme. Built in at the bottom, completely free at the top. This is a flagpole. It buckles into a quarter wave: vertical at the base, and leaning out at the top with no moment there at all. There is only one point of zero moment on the column, and it is the free end itself. So where is the other one? It is not on the column. Reflect the shape through the fixed base. The reflected curve meets the real one with the same slope, so the two together make one smooth curve, and it is exactly the pinned half sine wave, of total length two L. So the effective length is two L, the factor K is two, and the critical load is a quarter of the pinned value. A flagpole is four times weaker than the same member pinned at both ends. One case is left, and it is the only one that is not a round number. Pinned at one end and built in at the other. The pin gives zero moment, so one of the two points is the pin itself. The other one is the single inflection point up here. Where it sits comes out of a transcendental equation rather than a fraction, and it lands at about seven tenths of the length. So K is roughly zero point seven. Here they all are together. In every row, K is nothing more than the fraction of the column that lies between adjacent points of zero moment, and the critical load is pi squared E I over that length squared. One warning worth carrying. These are the idealised values. A real connection is never a perfect pin and never perfectly fixed, so design codes pull these numbers back toward one. But the geometry is the reason they are what they are. We have a critical load. To compare columns of different sizes we want a critical stress, so divide P critical by the cross sectional area. The cross section now appears twice in that expression, once as I and once as A. Those two are not independent, and the length that ties them together has a name. Define r, the radius of gyration, as the square root of I over A. It is the distance from the bending axis at which you could put the whole area and get the same second moment. For this rectangle, it is here. Substitute I equals A r squared. The area cancels, top and bottom, and what is left is pi squared E over the square of K L divided by r. That group is the slenderness ratio. Effective length over radius of gyration. It is dimensionless, it is the only geometry the answer needs, and it is what engineers actually mean when they call a column slender. So carry those two results across, and plot the critical stress against slenderness. It is a hyperbola: double the slenderness, and the critical stress falls by a factor of four. At a slenderness of two hundred, this steel column buckles at about fifty megapascals. Bring it in to a hundred and forty, and it is around a hundred. At a hundred, about two hundred. Now draw in the yield strength. Two hundred and fifty megapascals for a common structural steel. And look at what the Euler curve does on the left of the picture. It climbs straight through it. That part of the curve is a lie. It predicts a buckling stress well above the stress at which the material gives way. The column would have yielded long before it got there, and the derivation assumed elastic behaviour throughout. The crossing point is where the two mechanisms trade places. Set pi squared E over lambda squared equal to sigma y and solve. Lambda critical is pi times the square root of E over sigma y, and for this steel that is about eighty nine. So above about eighty nine, elastic buckling governs and the strength of the material barely matters. Below it, the column is stocky enough to reach yield first, and now the strength is what matters and the stiffness barely does. But that crossing is a sharp corner, and real columns do not have corners. Two things round it off, and both of them make the column weaker rather than stronger. First, every real column is slightly crooked to start with. So it begins bending from the very first increment of load, instead of waiting for a threshold and then choosing to bend. Second, a rolled section cools unevenly and locks in residual stresses. Part of the cross section is already close to yield before you apply any load at all, so it softens early and the whole member follows. The consequence is this red curve. Real columns sit below both idealisations, and the gap is worst right around the crossing, in the intermediate range, which is where most practical columns actually live. So codes do not use either straight idealisation. They use a fitted curve of this shape, calibrated against tests, with Euler as the upper bound it approaches once the column is slender enough. So, to gather it up. A slender column fails because the straight shape stops being the only equilibrium available, at a load of pi squared E I over the effective length squared. The end conditions enter only through where the points of zero moment sit. And the slenderness ratio tells you which failure you are actually designing against, with Euler an upper bound that gets more honest the more slender the column is.","watch":{"version":1,"scenes":[{"title":"Two Ways for a Column to Fail","start":0,"end":144.641875,"objects":{"card":"a Title that says \"Mechanics of Materials — Euler Buckling: Why Slender Columns Fail by Instability\"","head_model":"a Heading that says \"The Case We Will Solve\"","head_two":"a Heading that says \"Same Steel, Same Section, Different Length\"","label_long":"a Tex [text] that says \"Buckles: bows while the material is still elastic\"","label_short":"a Tex [text] that says \"Yields: $P = A sigma_y$, and no $L$ appears in it\"","mode_note":"a Tex [text] that says \"The shape at that load: one half sine wave.\"","model":"a Figure (x_range=(-1.9, 1.7), y_range=(-0.8, 7.2), aspect=(3.6, 8.0))","model_base":"a Line [gray] drawn in model (start=(-0.85, 0.0), end=(0.85, 0.0))","model_buckled":"a ParametricCurve [yellow] drawn in model (function=<function>, t_range=(0.0, 5.6))","model_cap":"a Line [gray] drawn in model (start=(-0.85, 5.6), end=(0.85, 5.6))","model_dim":"a Line [gray] labelled \"L\" drawn in model (start=(-1.45, 0.0), end=(-1.45, 5.6))","model_load":"an Arrow [red] labelled \"P\" drawn in model (start=(0.0, 6.6), end=(0.0, 5.85))","model_note":"a Block [text] that says \"Pinned at both ends: free to rotate, held sideways. 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Make the same strut three metres long, from the same steel, with the same cross section, and it folds up under a small fraction of that. Nothing about the material changed.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[14.42,"card is hidden from the screen — left the board."]]},{"start":15.62,"say":"Here are two columns cut from one bar. The short one fails the way you expect. Push hard enough, the stress reaches the yield strength of the steel, and the material gives way.","live":null,"does":[[15.62,"head_two is shown on the screen, written out."],[15.62,"stocky is shown on the screen, written out."],[15.62,"stocky_base is shown on the screen, written out."],[15.62,"stocky_shaft is shown on the screen, written out."],[21.495,"stocky_load is shown on the screen, written out."],[25.918,"stocky moves to a new place on the board."],[25.918,"label_short is shown on the screen, written out."]]},{"start":27.5865,"say":"Now the long one. Push on it, and long before the stress anywhere in it comes near yield, it does this. It bows sideways.","live":["label_short","stocky","head_two","stocky_base","stocky_shaft","stocky_load"],"does":[[27.5865,"slender is shown on the screen, written out."],[27.5865,"slender_base is shown on the screen, written out."],[27.5865,"slender_axis is shown on the screen, written out."],[29.688,"slender_load is shown on the screen, written out."],[35.435,"slender_shaft is shown on the screen, drawn."],[35.736000000000004,"label_long is shown on the screen, written out."]]},{"start":37.1435,"say":"The material is still perfectly elastic. Unload it and it springs straight again. But as a structural member it has failed, because it will not take any more load than that.","live":["label_short","stocky","label_long","slender","head_two","stocky_base","stocky_shaft","stocky_load","slender_base","slender_axis","slender_load","slender_shaft"],"does":[[38.833000000000006,"slender_shaft is indicated — a transient flash."],[41.20100000000001,"slender_axis is indicated — a transient flash."]]},{"start":48.477000000000004,"say":"And the gap between the two is not small. Triple the length, and the load the column can take falls by a factor of nine. Nothing in the material noticed.","live":null,"does":[[51.34500000000001,"stocky_shaft is indicated — a transient flash."],[55.21100000000001,"slender_shaft is indicated — a transient flash."]]},{"start":58.539500000000004,"say":"This is buckling, and it is a failure of stability rather than a failure of strength. The question is what load it happens at, and why the length matters so much when the material does not.","live":null,"does":[[70.6485,"head_two is hidden from the screen — left the board."],[70.6485,"label_long is hidden from the screen — left the board."],[70.6485,"label_short is hidden from the screen — left the board."],[70.6485,"slender is hidden from the screen — left the board."],[70.6485,"slender_base is hidden from the screen — slender left the board."],[70.6485,"slender_axis is hidden from the screen — slender left the board."],[70.6485,"slender_load is hidden from the screen — slender left the board."],[70.6485,"slender_shaft is hidden from the screen — slender left the board."],[70.6485,"stocky is hidden from the screen — left the board."],[70.6485,"stocky_base is hidden from the screen — stocky left the board."],[70.6485,"stocky_shaft is hidden from the screen — stocky left the board."],[70.6485,"stocky_load is hidden from the screen — stocky left the board."]]},{"start":71.8485,"say":"Here is the case we will solve. A straight column of length L, pinned at both ends, so each end is free to rotate but held against sideways movement. It carries an axial load P through the centroid.","live":[],"does":[[71.8485,"head_model is shown on the screen, written out."],[71.8485,"model is shown on the screen, written out."],[71.8485,"model_base is shown on the screen, written out."],[71.8485,"model_cap is shown on the screen, written out."],[71.8485,"model_shaft is shown on the screen, written out."],[75.598,"model_dim is shown on the screen, written out."],[76.539,"model_pin_b is shown on the screen, written out."],[76.539,"model_pin_t is shown on the screen, written out."],[82.947,"model_load is shown on the screen, written out."],[84.665,"model moves to a new place on the board."],[84.665,"model_note is shown on the screen, written out."]]},{"start":86.2175,"say":"The material is linearly elastic with Young's modulus E, and the cross section has a second moment of area I about the axis it bends about. Those are the only four quantities the answer can depend on.","live":["model_note","model","head_model","model_base","model_cap","model_shaft","model_dim","model_pin_b","model_pin_t","model_load"],"does":[[89.29400000000001,"model_note (the \"Linearly elastic material.\" part) is emphasized."],[91.93,"model_note (the \"Bending about one principal axis.\" part) is emphasized."],[91.93,"model_note (the \"Linearly elastic material.\" part) is no longer emphasized."],[96.36500000000001,"model_note (the \"Bending about one principal axis.\" part) is no longer emphasized."]]},{"start":99.7165,"say":"The answer we are heading for is this. The critical load is pi squared E I over L squared, and the shape the column takes at that load is a single smooth half sine wave, zero at both pins and largest at midheight.","live":null,"does":[[102.863,"p_cr is shown on the screen, written out."],[107.344,"model_buckled is shown on the screen, drawn."],[110.432,"mode_note is shown on the screen, written out."]]},{"start":115.3865,"say":"I want both of those to fall out of the mechanics rather than be handed to you. And notice already what that formula does not contain. There is no yield stress in it anywhere.","live":["model_note","p_cr","mode_note","model","head_model","model_base","model_cap","model_shaft","model_dim","model_pin_b","model_pin_t","model_load","model_buckled"],"does":[[121.78299999999999,"p_cr (the \"frac(pi^2 E I, L^2)\" part) is emphasized."],[126.4155,"p_cr (the \"frac(pi^2 E I, L^2)\" part) is no longer emphasized."]]},{"start":127.0155,"say":"The load at which a slender column buckles has nothing to do with how strong the material is. Only with how stiff it is, and with how the length and the cross section combine. So let the column bend, write equilibrium for the bent shape, and see what the mathematics demands.","live":null,"does":[[138.277,"model_buckled is indicated — a transient flash."],[143.60020833333334,"head_model is hidden from the screen — left the board."],[143.60020833333334,"mode_note is hidden from the screen — left the board."],[143.60020833333334,"model is hidden from the screen — left the board."],[143.60020833333334,"model_base is hidden from the screen — model left the board."],[143.60020833333334,"model_cap is hidden from the screen — model left the board."],[143.60020833333334,"model_shaft is hidden from the screen — model left the board."],[143.60020833333334,"model_dim is hidden from the screen — model left the board."],[143.60020833333334,"model_pin_b is hidden from the screen — model left the board."],[143.60020833333334,"model_pin_t is hidden from the screen — model left the board."],[143.60020833333334,"model_load is hidden from the screen — model left the board."],[143.60020833333334,"model_buckled is hidden from the screen — model left the board."],[143.60020833333334,"model_note is hidden from the screen — left the board."],[143.60020833333334,"p_cr is hidden from the screen — left the board."]]}]},{"title":"Equilibrium of the Bent Column","start":144.641875,"end":341.00456249999996,"objects":{"axis":"a Line [gray] drawn in bent (end=(0.0, 5.6), dashed=True)","bent":"a Figure (x_range=(-1.9, 2.4), y_range=(-1.1, 7.1), aspect=(4.3, 8.2))","cut":"a VariableNumber (initial_value=3.4)","cut_line":"a Line [gray] drawn in bent (start=(-1.25, <VariableNumber cut = 2.8>), end=(1.25, <VariableNumber cut = 2.8>), dashed=True)","cut_point":"a Point [yellow] drawn in bent (location=((0.72 * sin(((3.141592653589793 * cut) / 5.6))), <VariableNumb…)","head_bent":"a Heading that says \"Suppose It Is Not Straight\"","lever":"a Line [red] labelled \"v(x)\" drawn in bent (start=(0.0, <VariableNumber cut = 2.8>), end=((0.72 * sin(((3.141592653589793 * cut) / 5.6))), <VariableNumb…)","load_base":"an Arrow [red] labelled \"P\" drawn in bent (start=(0.0, -1.0), end=(0.0, -0.25))","load_top":"an Arrow [red] labelled \"P\" drawn in bent (start=(0.0, 6.6), end=(0.0, 5.85))","moment_arc":"a CurvedArrow [yellow] labelled \"M(x)\" drawn in bent (start=(((0.72 * sin(((3.141592653589793 * cut) / 5.6))) + 0.6), (cut …, end=(((0.72 * sin(((3.141592653589793 * cut) / 5.6))) + 0.6), (cut …)","pin_b":"a Circle [gray] drawn in bent (radius=0.13)","pin_t":"a Circle [gray] drawn in bent (center=(0.0, 5.6), radius=0.13)","shaft":"a ParametricCurve [blue] drawn in bent (function=<function>, t_range=(0.0, 5.6))","work":"a Derivation [text] that says \"$M(x) + P thin v(x) &= 0 \\ M(x) &= - P thin v(x) \\ E I frac(dif^2 v, dif x^2) &= M(x) \\ E I v'' + P thin v &= 0 \\ v'' + k^2 v &= 0, quad k^2 = frac(P, E I)$\"","x_dim":"a Line [gray] labelled \"x\" drawn in bent (start=(-1.55, 0.0), end=(-1.55, <VariableNumber cut = 2.8>))"},"beats":[{"start":144.641875,"say":"Suppose the column is not straight. Not because something pushed it out of line, but suppose it simply is, by some small amount, and ask whether that bent shape can stand in equilibrium.","live":[],"does":[[144.641875,"head_bent is shown on the screen, written out."],[144.641875,"bent is shown on the screen, written out."],[145.976875,"axis is shown on the screen, written out."],[145.976875,"pin_b is shown on the screen, written out."],[145.976875,"pin_t is shown on the screen, written out."],[153.395875,"shaft is shown on the screen, drawn."],[154.649875,"load_top is shown on the screen, written out."],[154.649875,"load_base is shown on the screen, written out."]]},{"start":156.491875,"say":"The dashed grey line is where the axis used to be. Take a height x above the bottom pin, and call the sideways movement of the axis at that station v of x.","live":["bent","head_bent","axis","pin_b","pin_t","shaft","load_top","load_base"],"does":[[157.095875,"axis is indicated — a transient flash."],[161.089875,"cut_line is shown on the screen, written out."],[161.089875,"x_dim is shown on the screen, written out."],[163.538875,"cut_point is shown on the screen, written out."],[163.538875,"lever is shown on the screen, written out."]]},{"start":167.842875,"say":"Now cut the column there, and keep the piece below the cut. At the bottom, the axial load P acts along its original line of action, because that pin has not moved sideways at all.","live":["bent","head_bent","axis","pin_b","pin_t","shaft","load_top","load_base","cut_line","x_dim","cut_point","lever"],"does":[[173.833875,"load_base is indicated — a transient flash."],[178.221875,"pin_b is indicated — a transient flash."]]},{"start":181.074375,"say":"And there is no sideways reaction at the pin either, because there is no sideways load anywhere on the column. So the base hands the free body exactly one force, P, at zero offset from the original axis.","live":null,"does":[[194.065875,"axis is indicated — a transient flash."]]},{"start":195.954375,"say":"That is the whole free body. One axial force at the base, and the internal shear and moment at the cut. Nothing else is acting on the piece.","live":null,"does":[]},{"start":206.538875,"say":"Take moments about the cut section. The load P is acting a distance v of x away from that section, so it applies a moment of P times v of x. The internal bending moment has to balance it exactly.","live":null,"does":[[210.915875,"lever is emphasized."],[217.60287499999998,"moment_arc is shown on the screen, written out."],[219.146875,"bent moves to a new place on the board."],[219.146875,"work is shown on the screen, written out."],[219.146875,"lever is no longer emphasized."]]},{"start":221.349875,"say":"So M of x plus P v of x is zero, and the bending moment at any section is minus the axial load, times how far that section has strayed from the line of action.","live":["bent","head_bent","axis","pin_b","pin_t","shaft","load_top","load_base","cut_line","x_dim","cut_point","lever","moment_arc"],"does":[[225.551875,"work is shown on the screen, written out."],[227.30587500000001,"work (the \"- P thin v(x)\" part) is emphasized."],[232.228375,"work (the \"- P thin v(x)\" part) is no longer emphasized."]]},{"start":232.828375,"say":"Watch what that means as the cut moves. Near the ends the deflection is small, so the moment is small. At midheight the deflection is largest, and so is the bending moment.","live":null,"does":[[235.811875,"cut_line is redrawn as the numbers it depends on change."],[235.811875,"x_dim is redrawn as the numbers it depends on change."],[235.811875,"cut_point is redrawn as the numbers it depends on change."],[235.811875,"lever is redrawn as the numbers it depends on change."],[235.811875,"moment_arc is redrawn as the numbers it depends on change."],[235.811875,"cut ticks to 0.65."],[240.56087499999998,"cut_line is redrawn as the numbers it depends on change."],[240.56087499999998,"x_dim is redrawn as the numbers it depends on change."],[240.56087499999998,"cut_point is redrawn as the numbers it depends on change."],[240.56087499999998,"lever is redrawn as the numbers it depends on change."],[240.56087499999998,"moment_arc is redrawn as the numbers it depends on change."],[240.56087499999998,"cut ticks to 2.8."]]},{"start":245.177875,"say":"So the bending moment along the column has the same shape as the deflection itself, scaled by P. That is unusual. In ordinary beam bending you are handed the loading and you go and find the moment. Here the moment depends on the answer.","live":null,"does":[[247.569875,"shaft is indicated — a transient flash."],[249.93787500000002,"moment_arc is indicated — a transient flash."]]},{"start":261.253875,"say":"And there is the loop that makes buckling what it is. Deflection produces bending moment. Bending moment produces curvature. Curvature produces more deflection. The load P is the gain around that loop.","live":null,"does":[[264.852875,"lever is indicated — a transient flash."],[265.909875,"moment_arc is indicated — a transient flash."],[268.927875,"shaft is indicated — a transient flash."],[274.604875,"load_top is indicated — a transient flash."]]},{"start":276.748875,"say":"Now bring in the bending relation you already have. For small deflections, E I times the second derivative of the deflection equals the bending moment at that section.","live":null,"does":[[282.85587499999997,"work is shown on the screen, written out."]]},{"start":288.436875,"say":"That is the small deflection form, and it is the honest limit of what we are doing here. As long as the bow stays small compared with the length, the curvature is well approximated by v double prime.","live":null,"does":[]},{"start":301.273375,"say":"Substitute the moment we just found. E I v double prime equals minus P v. Move everything to one side, and there is the governing equation.","live":null,"does":[[308.656875,"work (the \"E I v''\" part) is emphasized."],[310.247875,"work is shown on the screen, written out."],[311.565375,"work (the \"E I v''\" part) is no longer emphasized."]]},{"start":312.165375,"say":"Divide through by E I, and give the group P over E I a name. Call it k squared. Then the equation reads v double prime plus k squared v equals zero, and k carries units of one over length.","live":null,"does":[[317.488875,"work is shown on the screen, written out."],[318.138875,"work (the \"k^2 = frac(P, E I)\" part) is emphasized."],[326.18487500000003,"work (the \"k^2 = frac(P, E I)\" part) is no longer emphasized."]]},{"start":328.700375,"say":"That is a second order linear equation with constant coefficients, and you already know every function that satisfies it. Solving it is what turns this into a statement about the load.","live":null,"does":[[329.558875,"A box is drawn around work."],[339.9628958333333,"bent is hidden from the screen — left the board."],[339.9628958333333,"axis is hidden from the screen — bent left the board."],[339.9628958333333,"pin_b is hidden from the screen — bent left the board."],[339.9628958333333,"pin_t is hidden from the screen — bent left the board."],[339.9628958333333,"shaft is hidden from the screen — bent left the board."],[339.9628958333333,"load_top is hidden from the screen — bent left the board."],[339.9628958333333,"load_base is hidden from the screen — bent left the board."],[339.9628958333333,"cut_line is hidden from the screen — bent left the board."],[339.9628958333333,"x_dim is hidden from the screen — bent left the board."],[339.9628958333333,"cut_point is hidden from the screen — bent left the board."],[339.9628958333333,"lever is hidden from the screen — bent left the board."],[339.9628958333333,"moment_arc is hidden from the screen — bent left the board."],[339.9628958333333,"head_bent is hidden from the screen — left the board."],[339.9628958333333,"work is hidden from the screen — left the board."]]}]},{"title":"The Critical Load and the Mode Shape","start":341.00456249999996,"end":593.553375,"objects":{"bif_note":"a Panel that says \"Below the critical load the straight shape is the only equilibrium. At the critical load a whole family of bent equilibria appears, and the straight one stops being unique.\"","brace_mark":"a Point [green] labelled \"brace\" drawn in figure_2 (location=(0.0, 2.8), marker_radius=0.14)","branch_left":"a Line [red] drawn in diagram (start=(-1.02, 1.0), end=(0.0, 1.0))","branch_right":"a Line [red] labelled \"upright(\"buckled\")\" drawn in diagram (start=(0.0, 1.0), end=(1.02, 1.0))","cap1":"a Math [text] that says \"$P_1 = frac(pi^2 E I, L^2)$\"","cap2":"a Math [text] that says \"$P_2 = frac(4 pi^2 E I, L^2)$\"","cap3":"a Math [text] that says \"$P_3 = frac(9 pi^2 E I, L^2)$\"","col2_axis":"a Line [gray] drawn in figure_2 (end=(0.0, 5.6), dashed=True)","col2_pin_b":"a Circle [gray] drawn in figure_2 (radius=0.13)","col2_pin_t":"a Circle [gray] drawn in figure_2 (center=(0.0, 5.6), radius=0.13)","col3_axis":"a Line [gray] drawn in figure_3 (end=(0.0, 5.6), dashed=True)","col3_pin_b":"a Circle [gray] drawn in figure_3 (radius=0.13)","col3_pin_t":"a Circle [gray] drawn in figure_3 (center=(0.0, 5.6), radius=0.13)","col_axis":"a Line [gray] drawn in figure (end=(0.0, 5.6), dashed=True)","col_pin_b":"a Circle [gray] drawn in figure (radius=0.13)","col_pin_t":"a Circle [gray] drawn in figure (center=(0.0, 5.6), radius=0.13)","critical":"a Math [text] that says \"$P_upright(\"cr\") = frac(pi^2 E I, L^2)$\"","diagram":"an Axes (x_range=(-1.15, 1.15), y_range=(0.0, 2.3), include_ticks=False)","figure":"a Figure (x_range=(-1.7, 1.7), y_range=(-0.7, 6.7), aspect=(3.4, 7.4))","figure_2":"a Figure (x_range=(-1.7, 1.7), y_range=(-0.7, 6.7), aspect=(3.4, 7.4))","figure_3":"a Figure (x_range=(-1.7, 1.7), y_range=(-0.7, 6.7), aspect=(3.4, 7.4))","head_bif":"a Heading that says \"Where the Straight Shape Loses Its Monopoly\"","head_modes":"a Heading that says \"The Higher Modes Are Real Too\"","head_solve":"a Heading that says \"Solving It, and What the Answer Says\"","load":"a VariableNumber (initial_value=0.12)","mode1_big":"a ParametricCurve [blue] drawn in figure (function=<function>, t_range=(0.0, 5.6))","mode1_mid":"a ParametricCurve [blue] drawn in figure (function=<function>, t_range=(0.0, 5.6))","mode1_small":"a ParametricCurve [blue] drawn in figure (function=<function>, t_range=(0.0, 5.6))","mode2":"a ParametricCurve [blue] drawn in figure_2 (function=<function>, t_range=(0.0, 5.6))","mode3":"a ParametricCurve [blue] drawn in figure_3 (function=<function>, t_range=(0.0, 5.6))","pcr_tag":"a Math [text] that says \"$P_upright(\"cr\")$\" drawn in diagram","probe":"a Point [yellow] drawn in diagram (location=(0.0, <VariableNumber load = 1.0>), marker_radius=0.05)","straight_tag":"a Math [text] that says \"$upright(\"straight\")$\" drawn in diagram","trivial":"a Line [blue] drawn in diagram (end=(0.0, 2.2))","work":"a Derivation [text] that says \"$v(x) &= A sin(k x) + B cos(k x) \\ v(0) = 0 quad &arrow.r quad B = 0 \\ v(L) = 0 quad &arrow.r quad A sin(k L) = 0 \\ k L &= n pi, quad n = 1, 2, 3, dots \\ P_n &= frac(n^2 pi^2 E I, L^2) \\ P_upright(\"cr\") &= frac(pi^2 E I, L^2) \\ v(x) &= A si…$\""},"beats":[{"start":341.00456249999996,"say":"v double prime plus k squared v equals zero. Every solution of that is a combination of a sine and a cosine of k x, so write the general one down with two constants in it.","live":[],"does":[[341.00456249999996,"head_solve is shown on the screen, written out."],[341.00456249999996,"figure is shown on the screen, written out."],[341.00456249999996,"col_axis is shown on the screen, written out."],[341.00456249999996,"col_pin_b is shown on the screen, written out."],[341.00456249999996,"col_pin_t is shown on the screen, written out."],[350.83856249999997,"figure moves to a new place on the board."],[350.83856249999997,"work is shown on the screen, written out."]]},{"start":353.77206249999995,"say":"Now the end conditions. The bottom is pinned, so it cannot move sideways there. v of zero is zero. Put x equal to zero into the general solution: the sine term vanishes on its own, and what is left is B.","live":["figure","head_solve","col_axis","col_pin_b","col_pin_t"],"does":[[355.6525625,"col_pin_b is indicated — a transient flash."],[367.47156249999995,"work is shown on the screen, written out."]]},{"start":368.99506249999996,"say":"So B is zero, and that leaves v equals A sine k x. The top is pinned too, so v of L is zero as well. Put x equal to L in, and you get A times sine of k L equals zero.","live":null,"does":[[374.0975625,"col_pin_t is indicated — a transient flash."],[380.76156249999997,"work is shown on the screen, written out."]]},{"start":383.92756249999996,"say":"Look hard at that line, because everything is in it. A product of two things is zero, so at least one of them has to be zero, and there are exactly two ways for that to happen.","live":null,"does":[[388.0375625,"work (the \"A sin(k L) = 0\" part) is emphasized."],[393.6795625,"work (the \"A sin(k L) = 0\" part) is no longer emphasized."]]},{"start":396.34606249999996,"say":"The first way is that A is zero. That gives v identically zero, the perfectly straight column. And notice it is available at every value of P. The straight configuration is always an equilibrium, and it never stops being one.","live":null,"does":[[397.83256249999994,"work (the \"A\" part) is emphasized."],[401.8955625,"col_axis is emphasized."],[409.7325625,"col_axis is no longer emphasized."],[409.7325625,"work (the \"A\" part) is no longer emphasized."]]},{"start":411.91106249999996,"say":"The second way is that the sine itself is zero. Sine of k L vanishes when k L is a whole number of pi. And that is a condition on the load, because k squared is P over E I.","live":null,"does":[[413.47856249999995,"work (the \"sin(k L)\" part) is emphasized."],[418.25056249999994,"work is shown on the screen, written out."],[420.69956249999996,"work (the \"sin(k L)\" part) is no longer emphasized."]]},{"start":425.25306249999994,"say":"Square both sides, substitute for k squared, and P L squared over E I equals n squared pi squared. So P equals n squared pi squared E I over L squared.","live":null,"does":[[433.06056249999995,"work is shown on the screen, written out."]]},{"start":438.09606249999996,"say":"These are not loads the column happens to reach. They are the only loads at which anything other than the straight shape is possible at all. Below the smallest of them, A has to be zero.","live":null,"does":[[441.8115625,"work (the \"P_n\" part) is emphasized."],[446.16556249999996,"work (the \"P_n\" part) is no longer emphasized."]]},{"start":449.84206249999994,"say":"The smallest is n equal to one. Pi squared E I over L squared. That is the Euler critical load, and it came out of demanding that a bent shape can stand in equilibrium, not out of any strength calculation.","live":null,"does":[[452.8715625,"work is shown on the screen, written out."],[457.3185625,"work (the \"frac(pi^2 E I, L^2)\" part) is emphasized."],[463.88956249999995,"work (the \"frac(pi^2 E I, L^2)\" part) is no longer emphasized."]]},{"start":466.0450625,"say":"And the shape that goes with it is A sine of pi x over L. One half of a sine wave. Zero at both pins, largest at midheight, and no point of inflection anywhere in between.","live":null,"does":[[468.27456249999994,"work is shown on the screen, written out."],[471.22356249999996,"mode1_mid is shown on the screen, drawn."]]},{"start":479.56706249999996,"say":"Now the part that catches people out. A is still undetermined. The mathematics has fixed the shape and said nothing whatever about how big it is.","live":["figure","head_solve","col_axis","col_pin_b","col_pin_t","mode1_mid"],"does":[[483.39856249999997,"work (the \"A\" part) is emphasized."],[487.7875624999999,"work (the \"A\" part) is no longer emphasized."]]},{"start":489.59456249999994,"say":"Here is the same mode at three different amplitudes. Every one of them satisfies equilibrium exactly, and every one of them does it at exactly the same load. That is the signature of the thing we are looking at.","live":null,"does":[[491.10356249999995,"mode1_small is shown on the screen, written out."],[491.68456249999997,"mode1_big is shown on the screen, written out."],[500.5085624999999,"mode1_small is hidden from the screen."],[500.5085624999999,"mode1_big is hidden from the screen."],[502.63306249999994,"figure moves to a new place on the board."],[502.63306249999994,"head_solve is hidden from the screen — left the board."],[502.63306249999994,"work is hidden from the screen — left the board."]]},{"start":503.23306249999996,"say":"The higher values of n are genuine solutions too. n equal to two is a full sine wave, with a stationary point at midheight, and it needs four times the load. n equal to three needs nine times.","live":["figure","col_axis","col_pin_b","col_pin_t","mode1_mid"],"does":[[503.23306249999996,"head_modes is shown on the screen, written out."],[503.23306249999996,"cap1 is shown on the screen, written out."],[507.8885624999999,"figure_2 is shown on the screen, written out."],[507.8885624999999,"col2_axis is shown on the screen, written out."],[507.8885624999999,"col2_pin_b is shown on the screen, written out."],[507.8885624999999,"col2_pin_t is shown on the screen, written out."],[508.56256249999996,"mode2 is shown on the screen, drawn."],[512.7065625,"cap2 is shown on the screen, written out."],[515.1215625,"figure_3 is shown on the screen, written out."],[515.1215625,"col3_axis is shown on the screen, written out."],[515.1215625,"col3_pin_b is shown on the screen, written out."],[515.1215625,"col3_pin_t is shown on the screen, written out."],[515.4215624999999,"mode3 is shown on the screen, drawn."],[516.0505625,"cap3 is shown on the screen, written out."]]},{"start":517.8580625,"say":"In a bare column they never happen, because the column has already gone at n equal to one. But brace it sideways at midheight, and you have forbidden the first mode. The second one is what you get instead.","live":["figure","col_axis","col_pin_b","col_pin_t","mode1_mid","cap1","figure_2","cap2","figure_3","cap3","head_modes","col2_axis","col2_pin_b","col2_pin_t","mode2","col3_axis","col3_pin_b","col3_pin_t","mode3"],"does":[[524.0695625,"brace_mark is shown on the screen, written out."],[528.7245624999999,"mode2 is indicated — a transient flash."],[529.6885625,"cap2 is indicated — a transient flash."],[530.7455625,"cap1 is hidden from the screen — left the board."],[530.7455625,"cap2 is hidden from the screen — left the board."],[530.7455625,"cap3 is hidden from the screen — left the board."],[530.7455625,"figure is hidden from the screen — left the board."],[530.7455625,"col_axis is hidden from the screen — figure left the board."],[530.7455625,"col_pin_b is hidden from the screen — figure left the board."],[530.7455625,"col_pin_t is hidden from the screen — figure left the board."],[530.7455625,"mode1_mid is hidden from the screen — figure left the board."],[530.7455625,"figure_2 is hidden from the screen — left the board."],[530.7455625,"col2_axis is hidden from the screen — figure_2 left the board."],[530.7455625,"col2_pin_b is hidden from the screen — figure_2 left the board."],[530.7455625,"col2_pin_t is hidden from the screen — figure_2 left the board."],[530.7455625,"mode2 is hidden from the screen — figure_2 left the board."],[530.7455625,"brace_mark is hidden from the screen — figure_2 left the board."],[530.7455625,"figure_3 is hidden from the screen — left the board."],[530.7455625,"col3_axis is hidden from the screen — figure_3 left the board."],[530.7455625,"col3_pin_b is hidden from the screen — figure_3 left the board."],[530.7455625,"col3_pin_t is hidden from the screen — figure_3 left the board."],[530.7455625,"mode3 is hidden from the screen — figure_3 left the board."],[530.7455625,"head_modes is hidden from the screen — left the board."]]},{"start":531.3455624999999,"say":"Put all of that on one picture. Load runs up the vertical axis. The amplitude of the bow runs across the horizontal one, so the whole vertical axis is the straight column.","live":[],"does":[[531.3455624999999,"head_bif is shown on the screen, written out."],[531.3455624999999,"diagram is shown on the screen, written out."],[534.9785625,"trivial is shown on the screen, written out."],[542.1425624999999,"straight_tag is shown on the screen, written out."]]},{"start":543.6940625,"say":"Load it up from nothing. Below the critical load there is exactly one equilibrium at every value of P, and it sits on that axis at zero amplitude.","live":["diagram","head_bif","trivial","straight_tag"],"does":[[543.8565625,"probe is shown on the screen, written out."],[545.9235625,"probe is redrawn as the numbers it depends on change."],[545.9235625,"load ticks to 0.93."]]},{"start":554.7085625,"say":"At the critical load, a second branch appears. Every amplitude on that horizontal line is an equilibrium, at the one load, and the straight solution has stopped being unique.","live":["diagram","head_bif","trivial","straight_tag","probe"],"does":[[555.3815625,"probe is redrawn as the numbers it depends on change."],[555.3815625,"load ticks to 1.0."],[556.5775625,"pcr_tag is shown on the screen, written out."],[559.6895625,"branch_right is shown on the screen, drawn."],[559.8895625,"branch_left is shown on the screen, drawn."]]},{"start":566.6515625,"say":"That splitting is called a bifurcation, and it is exactly what we mean when we say the column has buckled. Nothing broke. No stress reached any limit. The column simply ran out of reasons to stay straight.","live":["diagram","head_bif","trivial","straight_tag","probe","pcr_tag","branch_right","branch_left"],"does":[[568.2655625,"diagram moves to a new place on the board."],[568.2655625,"bif_note is shown on the screen, written out."],[572.4685625,"critical is shown on the screen, written out."],[578.8075625,"trivial is indicated — a transient flash."]]},{"start":580.5215625,"say":"Now, that whole derivation used one particular pair of end conditions. Change them and the numbers change. What is worth seeing is that they change in a way you can read straight off the deflected shape.","live":["bif_note","critical","diagram","head_bif","trivial","straight_tag","probe","pcr_tag","branch_right","branch_left"],"does":[[580.5215625,"A box is drawn around critical."],[592.5117083333333,"bif_note is hidden from the screen — left the board."],[592.5117083333333,"critical is hidden from the screen — left the board."],[592.5117083333333,"diagram is hidden from the screen — left the board."],[592.5117083333333,"trivial is hidden from the screen — diagram left the board."],[592.5117083333333,"straight_tag is hidden from the screen — diagram left the board."],[592.5117083333333,"probe is hidden from the screen — diagram left the board."],[592.5117083333333,"pcr_tag is hidden from the screen — diagram left the board."],[592.5117083333333,"branch_right is hidden from the screen — diagram left the board."],[592.5117083333333,"branch_left is hidden from the screen — diagram left the board."],[592.5117083333333,"head_bif is hidden from the screen — left the board."]]}]},{"title":"Where the Bending Moment Vanishes","start":593.553375,"end":815.0678958333333,"objects":{"fc":"a Figure (x_range=(-2.3, 2.0), y_range=(-4.9, 5.0), aspect=(4.3, 9.9))","fc_base1":"a Line [gray] drawn in fc (start=(-0.9, 0.0), end=(0.9, 0.0))","fc_curve":"a ParametricCurve [blue] drawn in fc (function=<function>, t_range=(0.0, 4.0))","fc_full":"a Line [gray] labelled \"L\" drawn in fc (start=(1.55, 0.0), end=(1.55, 4.0))","fc_le":"a Line [green] labelled \"L_e = 2 L\" drawn in fc (start=(-1.85, -4.0), end=(-1.85, 4.0))","fc_len":"a Math [text] that says \"$L_e = 2 L, quad K = 2$\"","fc_load":"a Math [text] that says \"$P_upright(\"cr\") = frac(pi^2 E I, 4 L^2)$\"","fc_mirror":"a ParametricCurve [yellow] drawn in fc (function=<function>, t_range=(-4.0, 0.0))","fc_note":"a Tex [text] that says \"Reflect through the base: one smooth half wave of length $2L$.\"","fc_zero_m":"a Point [red] drawn in fc (location=(-0.62, -4.0), marker_radius=0.11)","fc_zero_t":"a Point [red] drawn in fc (location=(0.62, 4.0), marker_radius=0.11)","ff_base1":"a Line [gray] drawn in figure_2 (start=(-0.9, 0.0), end=(0.9, 0.0))","ff_base2":"a Line [gray] drawn in figure_2 (start=(-0.9, -0.18), end=(0.9, -0.18))","ff_cap1":"a Line [gray] drawn in figure_2 (start=(-0.9, 5.0), end=(0.9, 5.0))","ff_cap2":"a Line [gray] drawn in figure_2 (start=(-0.9, 5.18), end=(0.9, 5.18))","ff_curve":"a ParametricCurve [blue] drawn in figure_2 (function=<function>, t_range=(0.0, 5.0))","ff_full":"a Line [gray] labelled \"L\" drawn in figure_2 (start=(1.25, 0.0), end=(1.25, 5.0))","ff_inf_hi":"a Point [red] drawn in figure_2 (location=(0.31, 3.75), marker_radius=0.11)","ff_inf_lo":"a Point [red] drawn in figure_2 (location=(0.31, 1.25), marker_radius=0.11)","ff_infl":"a Math [text] that says \"$v'' = 0 quad arrow.r quad x = frac(L, 4), thin frac(3 L, 4)$\"","ff_le":"a Line [green] labelled \"L_e\" drawn in figure_2 (start=(-1.5, 1.25), end=(-1.5, 3.75))","ff_len":"a Math [text] that says \"$L_e = frac(L, 2), quad K = 0.5$\"","ff_load":"a Math [text] that says \"$P_upright(\"cr\") = frac(4 pi^2 E I, L^2)$\"","ff_shape":"a Math [text] that says \"$v(x) = frac(A, 2) (1 - cos(frac(2 pi x, L)))$\"","figure":"a Figure (x_range=(-1.95, 1.6), y_range=(-0.8, 5.9), aspect=(3.55, 6.7))","figure_2":"a Figure (x_range=(-1.95, 1.6), y_range=(-0.8, 5.9), aspect=(3.55, 6.7))","figure_3":"a Figure (x_range=(-1.95, 1.6), y_range=(-0.8, 5.9), aspect=(3.55, 6.7))","fp_base":"a Line [gray] drawn in figure_3 (start=(-0.8, 0.0), end=(0.8, 0.0))","fp_cap1":"a Line [gray] drawn in figure_3 (start=(-0.9, 5.0), end=(0.9, 5.0))","fp_cap2":"a Line [gray] drawn in figure_3 (start=(-0.9, 5.18), end=(0.9, 5.18))","fp_curve":"a ParametricCurve [blue] drawn in figure_3 (function=<function>, t_range=(0.0, 5.0))","fp_inf":"a Point [red] drawn in figure_3 (location=(0.30993184316599487, 3.49605), marker_radius=0.11)","fp_le":"a Line [green] labelled \"L_e\" drawn in figure_3 (start=(-1.5, 0.0), end=(-1.5, 3.49605))","fp_pin":"a Circle [gray] drawn in figure_3 (radius=0.12)","fp_zero_b":"a Point [red] drawn in figure_3 (marker_radius=0.11)","general_k":"a Math [text] that says \"$P_upright(\"cr\") = frac(pi^2 E I, (K L)^2)$\"","general_le":"a Math [text] that says \"$P_upright(\"cr\") = frac(pi^2 E I, L_e^2)$\"","head_all":"a Heading that says \"Four Cases, One Rule\"","head_eff":"a Heading that says \"The Statement That Replaces the Table\"","head_fc":"a Heading that says \"Built In at the Bottom, Free at the Top\"","head_ff":"a Heading that says \"Built In at Both Ends\"","k_table":"a Table [text] that says \"End conditions Zero moment at $K$ Pinned, pinned both ends 1 Fixed, fixed the quarter points 0.5 Fixed, free the top and its mirror 2 Fixed, pinned the pin and $0.7 L$ 0.7\" (rows=(('End conditions', 'Zero moment at', '$K$'), ('Pinned, pinned'…, header=True)","pp_base":"a Line [gray] drawn in figure (start=(-0.8, 0.0), end=(0.8, 0.0))","pp_cap":"a Line [gray] drawn in figure (start=(-0.8, 5.0), end=(0.8, 5.0))","pp_curve":"a ParametricCurve [blue] drawn in figure (function=<function>, t_range=(0.0, 5.0))","pp_le":"a Line [green] labelled \"L_e = L\" drawn in figure (start=(-1.5, 0.0), end=(-1.5, 5.0))","pp_pin_b":"a Circle [gray] drawn in figure (radius=0.12)","pp_pin_t":"a Circle [gray] drawn in figure (center=(0.0, 5.0), radius=0.12)","pp_zero_b":"a Point [red] drawn in figure (marker_radius=0.11)","pp_zero_t":"a Point [red] drawn in figure (location=(0.0, 5.0), marker_radius=0.11)","principle":"a Panel that says \"The effective length is the distance between adjacent points of zero bending moment in the buckled shape. Between two such points the curve is the pin ended half sine wave.\""},"beats":[{"start":593.553375,"say":"Everything in that derivation came from two facts about the ends. They could not move sideways, and they could not carry bending moment, because a pin cannot. Change either one and the answer changes.","live":[],"does":[[593.553375,"head_eff is shown on the screen, written out."],[593.553375,"figure is shown on the screen, written out."],[593.553375,"pp_base is shown on the screen, written out."],[593.553375,"pp_cap is shown on the screen, written out."],[593.553375,"pp_curve is shown on the screen, written out."],[601.111375,"pp_pin_b is shown on the screen, written out."],[601.111375,"pp_pin_t is shown on the screen, written out."]]},{"start":605.4438749999999,"say":"But look at what the pinned answer really is. The moment is P times the deflection, and the deflection is zero at both pins. So the half wave runs between two points where the bending moment is zero.","live":["figure","head_eff","pp_base","pp_cap","pp_curve","pp_pin_b","pp_pin_t"],"does":[[612.311375,"pp_zero_b is shown on the screen, written out."],[612.311375,"pp_zero_t is shown on the screen, written out."]]},{"start":618.461375,"say":"So here is the statement that replaces the table. The critical load is pi squared E I divided by the square of the distance between adjacent points of zero bending moment. Call that distance the effective length.","live":["figure","head_eff","pp_base","pp_cap","pp_curve","pp_pin_b","pp_pin_t","pp_zero_b","pp_zero_t"],"does":[[621.990375,"figure moves to a new place on the board."],[621.990375,"general_le is shown on the screen, written out."],[630.373375,"pp_le is shown on the screen, written out."],[630.373375,"principle is shown on the screen, written out."]]},{"start":632.226875,"say":"The reason is simple. Between two such points, the deflected curve is a half sine wave with no moment at either end. That is precisely the pin ended problem we already solved, sitting inside a longer column, and it does not care what happens beyond those two points.","live":["principle","general_le","figure","head_eff","pp_base","pp_cap","pp_curve","pp_pin_b","pp_pin_t","pp_zero_b","pp_zero_t","pp_le"],"does":[[636.859375,"pp_curve is emphasized."],[645.845375,"pp_curve is no longer emphasized."]]},{"start":649.313375,"say":"And zero bending moment means zero curvature, so those points are the inflection points of the deflected shape. You can find them by looking at where the curve changes the way it bends.","live":null,"does":[[651.159375,"principle (the \"zero\" part) is emphasized."],[659.983375,"figure is hidden from the screen — left the board."],[659.983375,"pp_base is hidden from the screen — figure left the board."],[659.983375,"pp_cap is hidden from the screen — figure left the board."],[659.983375,"pp_curve is hidden from the screen — figure left the board."],[659.983375,"pp_pin_b is hidden from the screen — figure left the board."],[659.983375,"pp_pin_t is hidden from the screen — figure left the board."],[659.983375,"pp_zero_b is hidden from the screen — figure left the board."],[659.983375,"pp_zero_t is hidden from the screen — figure left the board."],[659.983375,"pp_le is hidden from the screen — figure left the board."],[659.983375,"general_le is hidden from the screen — left the board."],[659.983375,"head_eff is hidden from the screen — left the board."],[659.983375,"principle is hidden from the screen — left the board."],[659.983375,"principle (the \"zero\" part) is no longer emphasized."]]},{"start":661.183375,"say":"Take a column built in at both ends, so neither end can rotate. Load it until it buckles, and it takes this shape: vertical where it meets each support, bulging one way in the middle, and curling back near each end.","live":[],"does":[[661.183375,"head_ff is shown on the screen, written out."],[661.183375,"figure_2 is shown on the screen, written out."],[661.183375,"ff_base1 is shown on the screen, written out."],[661.183375,"ff_base2 is shown on the screen, written out."],[661.183375,"ff_cap1 is shown on the screen, written out."],[661.183375,"ff_cap2 is shown on the screen, written out."],[668.265375,"ff_curve is shown on the screen, drawn."],[671.1553749999999,"figure_2 moves to a new place on the board."],[671.1553749999999,"ff_shape is shown on the screen, written out."]]},{"start":675.041375,"say":"Find the inflection points. The curve bends one way near the middle and the other way close to each support. It changes over here, and again here, at a quarter of the length in from each end.","live":["ff_shape","figure_2","head_ff","ff_base1","ff_base2","ff_cap1","ff_cap2","ff_curve"],"does":[[682.541375,"ff_inf_lo is shown on the screen, written out."],[683.388375,"ff_inf_hi is shown on the screen, written out."],[684.608375,"ff_infl is shown on the screen, written out."]]},{"start":687.082875,"say":"So the distance between them is half the length of the column. That middle half is a half sine wave with zero moment at both ends, and it has no idea it is not a pinned column of length L over two.","live":["ff_shape","ff_infl","figure_2","head_ff","ff_base1","ff_base2","ff_cap1","ff_cap2","ff_curve","ff_inf_lo","ff_inf_hi"],"does":[[687.599375,"ff_le is shown on the screen, written out."],[687.599375,"ff_full is shown on the screen, written out."],[690.8043749999999,"ff_len is shown on the screen, written out."]]},{"start":699.077875,"say":"Put L over two in place of L. The critical load is four times what the pinned column would carry. The factor K is one half, and it is not a number from a table. It is where the curvature vanishes.","live":["ff_shape","ff_infl","ff_len","figure_2","head_ff","ff_base1","ff_base2","ff_cap1","ff_cap2","ff_curve","ff_inf_lo","ff_inf_hi","ff_le","ff_full"],"does":[[702.653375,"ff_load is shown on the screen, written out."],[710.281375,"ff_inf_lo is indicated — a transient flash."],[710.281375,"ff_inf_hi is indicated — a transient flash."],[711.744875,"ff_infl is hidden from the screen — left the board."],[711.744875,"ff_len is hidden from the screen — left the board."],[711.744875,"ff_load is hidden from the screen — left the board."],[711.744875,"ff_shape is hidden from the screen — left the board."],[711.744875,"figure_2 is hidden from the screen — left the board."],[711.744875,"ff_base1 is hidden from the screen — figure_2 left the board."],[711.744875,"ff_base2 is hidden from the screen — figure_2 left the board."],[711.744875,"ff_cap1 is hidden from the screen — figure_2 left the board."],[711.744875,"ff_cap2 is hidden from the screen — figure_2 left the board."],[711.744875,"ff_curve is hidden from the screen — figure_2 left the board."],[711.744875,"ff_inf_lo is hidden from the screen — figure_2 left the board."],[711.744875,"ff_inf_hi is hidden from the screen — figure_2 left the board."],[711.744875,"ff_le is hidden from the screen — figure_2 left the board."],[711.744875,"ff_full is hidden from the screen — figure_2 left the board."],[711.744875,"head_ff is hidden from the screen — left the board."]]},{"start":712.944875,"say":"Now the other extreme. Built in at the bottom, completely free at the top. This is a flagpole. It buckles into a quarter wave: vertical at the base, and leaning out at the top with no moment there at all.","live":[],"does":[[712.944875,"head_fc is shown on the screen, written out."],[712.944875,"fc is shown on the screen, written out."],[712.944875,"fc_base1 is shown on the screen, written out."],[720.177375,"fc_curve is shown on the screen, drawn."],[720.177375,"fc_full is shown on the screen, written out."],[723.974375,"fc_zero_t is shown on the screen, written out."]]},{"start":725.978875,"say":"There is only one point of zero moment on the column, and it is the free end itself. So where is the other one? It is not on the column.","live":["fc","head_fc","fc_base1","fc_curve","fc_full","fc_zero_t"],"does":[[729.566375,"fc_zero_t is indicated — a transient flash."]]},{"start":735.158875,"say":"Reflect the shape through the fixed base. The reflected curve meets the real one with the same slope, so the two together make one smooth curve, and it is exactly the pinned half sine wave, of total length two L.","live":null,"does":[[735.379375,"fc_mirror is shown on the screen, drawn."],[741.788375,"fc_zero_m is shown on the screen, written out."],[745.735375,"fc moves to a new place on the board."],[745.735375,"fc_le is shown on the screen, written out."],[745.735375,"fc_note is shown on the screen, written out."]]},{"start":747.879875,"say":"So the effective length is two L, the factor K is two, and the critical load is a quarter of the pinned value. A flagpole is four times weaker than the same member pinned at both ends.","live":["fc_note","fc","head_fc","fc_base1","fc_curve","fc_full","fc_zero_t","fc_mirror","fc_zero_m","fc_le"],"does":[[750.085375,"fc_len is shown on the screen, written out."],[752.674375,"fc_load is shown on the screen, written out."],[758.746875,"fc is hidden from the screen — left the board."],[758.746875,"fc_base1 is hidden from the screen — fc left the board."],[758.746875,"fc_curve is hidden from the screen — fc left the board."],[758.746875,"fc_full is hidden from the screen — fc left the board."],[758.746875,"fc_zero_t is hidden from the screen — fc left the board."],[758.746875,"fc_mirror is hidden from the screen — fc left the board."],[758.746875,"fc_zero_m is hidden from the screen — fc left the board."],[758.746875,"fc_le is hidden from the screen — fc left the board."],[758.746875,"fc_len is hidden from the screen — left the board."],[758.746875,"fc_load is hidden from the screen — left the board."],[758.746875,"fc_note is hidden from the screen — left the board."],[758.746875,"head_fc is hidden from the screen — left the board."]]},{"start":759.946875,"say":"One case is left, and it is the only one that is not a round number. Pinned at one end and built in at the other. The pin gives zero moment, so one of the two points is the pin itself.","live":[],"does":[[759.946875,"head_all is shown on the screen, written out."],[759.946875,"figure_3 is shown on the screen, written out."],[759.946875,"fp_base is shown on the screen, written out."],[759.946875,"fp_cap1 is shown on the screen, written out."],[759.946875,"fp_cap2 is shown on the screen, written out."],[764.556375,"fp_curve is shown on the screen, drawn."],[767.435375,"fp_pin is shown on the screen, written out."],[767.435375,"fp_zero_b is shown on the screen, written out."]]},{"start":772.401375,"say":"The other one is the single inflection point up here. Where it sits comes out of a transcendental equation rather than a fraction, and it lands at about seven tenths of the length. So K is roughly zero point seven.","live":["figure_3","head_all","fp_base","fp_cap1","fp_cap2","fp_curve","fp_pin","fp_zero_b"],"does":[[774.7693750000001,"fp_inf is shown on the screen, written out."],[780.899375,"fp_le is shown on the screen, written out."]]},{"start":785.713875,"say":"Here they all are together. In every row, K is nothing more than the fraction of the column that lies between adjacent points of zero moment, and the critical load is pi squared E I over that length squared.","live":["figure_3","head_all","fp_base","fp_cap1","fp_cap2","fp_curve","fp_pin","fp_zero_b","fp_inf","fp_le"],"does":[[785.713875,"figure_3 moves to a new place on the board."],[785.713875,"k_table is shown on the screen, written out."],[787.920375,"k_table is shown on the screen, written out."],[788.170375,"k_table is shown on the screen, written out."],[788.670375,"k_table is shown on the screen, written out."],[789.420375,"k_table is shown on the screen, written out."],[789.998375,"k_table (the \"column=3\" part) is emphasized."],[794.386375,"general_k is shown on the screen, written out."],[798.391875,"k_table (the \"column=3\" part) is no longer emphasized."]]},{"start":798.9918749999999,"say":"One warning worth carrying. These are the idealised values. A real connection is never a perfect pin and never perfectly fixed, so design codes pull these numbers back toward one. But the geometry is the reason they are what they are.","live":["general_k","figure_3","head_all","fp_base","fp_cap1","fp_cap2","fp_curve","fp_pin","fp_zero_b","fp_inf","fp_le"],"does":[[801.8943750000001,"A box is drawn around general_k."],[814.0262291666667,"figure_3 is hidden from the screen — left the board."],[814.0262291666667,"fp_base is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_cap1 is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_cap2 is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_curve is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_pin is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_zero_b is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_inf is hidden from the screen — figure_3 left the board."],[814.0262291666667,"fp_le is hidden from the screen — figure_3 left the board."],[814.0262291666667,"general_k is hidden from the screen — left the board."],[814.0262291666667,"head_all is hidden from the screen — left the board."],[814.0262291666667,"k_table is hidden from the screen — left the board."]]}]},{"title":"Slenderness, and Where Euler Stops","start":815.0678958333333,"end":1051.7873124999999,"objects":{"chart":"an Axes (x_range=(0.0, 215.0), y_range=(0.0, 520.0), x_ticks_every=50.0)","crossing":"a Point [red] labelled \"lambda_c\" drawn in chart (location=(88.85765876316732, 250.0))","cutoff":"a Line [gray] drawn in chart (start=(88.85765876316732, 0.0), end=(88.85765876316732, 250.0), dashed=True)","euler":"a FunctionPlot [blue] labelled \"upright(\"Euler\")\" drawn in chart (function=<function>, x_range=(64.0, 212.0))","head_curve":"a Heading that says \"Critical Stress Against Slenderness\"","head_real":"a Heading that says \"What Real Columns Actually Do\"","head_slend":"a Heading that says \"From a Critical Load to a Critical Stress\"","lam":"a VariableNumber (initial_value=200.0, format_spec='.0f')","lam_c_eq":"a Math [text] that says \"$lambda_c = pi sqrt(frac(E, sigma_y)) approx 89$\"","lam_def":"a Math [text] that says \"$lambda = frac(K L, r)$\"","point":"a Point [yellow] drawn in chart (location=(70.0, 402.84099596283096))","point_2":"a Point [yellow] drawn in chart (location=(88.85765876316732, 176.77669529663686))","probe":"a PlotPoint [yellow] labelled \"200\" drawn in chart (target='euler', x=<VariableNumber lam = 70.0>)","r_dim":"a Line [yellow] labelled \"r\" drawn in sec (start=(1.62, 0.0), end=(1.62, 0.5196152422706632))","r_dn":"a Line [yellow] drawn in sec (start=(-1.4, -0.5196152422706632), end=(1.4, -0.5196152422706632), dashed=True)","r_up":"a Line [yellow] drawn in sec (start=(-1.4, 0.5196152422706632), end=(1.4, 0.5196152422706632), dashed=True)","real":"a FunctionPlot [red] labelled \"upright(\"real columns\")\" drawn in chart (function=<function>, x_range=(10.0, 212.0))","reasons":"a Block [text] that says \"Every real column is slightly crooked, so it bends from the first increment of load. Rolled sections carry residual stresses, so part of the section yields early. Codes fit a smooth curve to test data, below both idealisations.\"","rect":"a Polygon [blue] drawn in sec (vertices=((-1.4, -0.9), (1.4, -0.9), (1.4, 0.9), (-1.4, 0.9)), fill_opacity=0.2)","sec":"a Figure (x_range=(-2.2, 2.2), y_range=(-2.2, 2.2), aspect=(1.0, 1.0))","sec_axis":"a Line [gray] drawn in sec (start=(-1.85, 0.0), end=(1.85, 0.0), dashed=True)","sigma_law":"a Math [text] that says \"$sigma_upright(\"cr\") = frac(pi^2 E, lambda^2)$\"","work":"a Derivation [text] that says \"$sigma_upright(\"cr\") &= frac(P_upright(\"cr\"), A) = frac(pi^2 E I, A (K L)^2) \\ r &= sqrt(frac(I, A)) quad arrow.r quad I = A r^2 \\ sigma_upright(\"cr\") &= frac(pi^2 E A r^2, A (K L)^2) \\ &= frac(pi^2 E, (frac(K L, r))^2) \\ lambda &= frac(K L…$\"","yield_line":"a Line [yellow] labelled \"sigma_y\" drawn in chart (start=(0.0, 250.0), end=(212.0, 250.0), dashed=True)"},"beats":[{"start":815.0678958333333,"say":"We have a critical load. To compare columns of different sizes we want a critical stress, so divide P critical by the cross sectional area.","live":[],"does":[[815.0678958333333,"head_slend is shown on the screen, written out."],[820.9428958333333,"work is shown on the screen, written out."],[822.2778958333333,"sec is shown on the screen, written out."],[822.2778958333333,"rect is shown on the screen, written out."],[822.2778958333333,"sec_axis is shown on the screen, written out."]]},{"start":824.5378958333333,"say":"The cross section now appears twice in that expression, once as I and once as A. Those two are not independent, and the length that ties them together has a name.","live":["sec","head_slend","rect","sec_axis"],"does":[[827.1728958333333,"work (the \"I\" part) is emphasized."],[829.2978958333333,"work (the \"A\" part) is emphasized."],[829.2978958333333,"work (the \"I\" part) is no longer emphasized."],[833.8028958333333,"work (the \"A\" part) is no longer emphasized."]]},{"start":836.2718958333332,"say":"Define r, the radius of gyration, as the square root of I over A. It is the distance from the bending axis at which you could put the whole area and get the same second moment. For this rectangle, it is here.","live":null,"does":[[836.6198958333333,"work is shown on the screen, written out."],[842.6918958333333,"r_up is shown on the screen, written out."],[842.6918958333333,"r_dn is shown on the screen, written out."],[850.4708958333333,"r_dim is shown on the screen, written out."]]},{"start":851.7443958333333,"say":"Substitute I equals A r squared. The area cancels, top and bottom, and what is left is pi squared E over the square of K L divided by r.","live":["sec","head_slend","rect","sec_axis","r_up","r_dn","r_dim"],"does":[[852.0928958333333,"work is shown on the screen, written out."],[855.8198958333332,"work (the \"A\" part) is slashed through — it cancels."],[855.8198958333332,"work (the \"A#2\" part) is slashed through — it cancels."],[857.3748958333333,"work is shown on the screen, written out."]]},{"start":862.8048958333333,"say":"That group is the slenderness ratio. Effective length over radius of gyration. It is dimensionless, it is the only geometry the answer needs, and it is what engineers actually mean when they call a column slender.","live":null,"does":[[864.0008958333333,"work is shown on the screen, written out."],[869.4688958333332,"work is shown on the screen, written out."],[871.0708958333333,"work (the \"frac(K L, r)\" part) is emphasized."],[876.1568958333332,"work (the \"frac(K L, r)\" part) is no longer emphasized."],[877.0853958333333,"head_slend is hidden from the screen — left the board."],[877.0853958333333,"sec is hidden from the screen — left the board."],[877.0853958333333,"rect is hidden from the screen — sec left the board."],[877.0853958333333,"sec_axis is hidden from the screen — sec left the board."],[877.0853958333333,"r_up is hidden from the screen — sec left the board."],[877.0853958333333,"r_dn is hidden from the screen — sec left the board."],[877.0853958333333,"r_dim is hidden from the screen — sec left the board."],[877.0853958333333,"work is hidden from the screen — left the board."]]},{"start":877.6853958333332,"say":"So carry those two results across, and plot the critical stress against slenderness. It is a hyperbola: double the slenderness, and the critical stress falls by a factor of four.","live":[],"does":[[877.6853958333332,"head_curve is shown on the screen, written out."],[878.8458958333333,"lam_def is shown on the screen, written out."],[879.1458958333333,"sigma_law is shown on the screen, written out."],[880.4718958333333,"chart is shown on the screen, written out."],[883.9898958333333,"euler is shown on the screen, drawn."]]},{"start":889.6633958333333,"say":"At a slenderness of two hundred, this steel column buckles at about fifty megapascals. Bring it in to a hundred and forty, and it is around a hundred. At a hundred, about two hundred.","live":["lam_def","sigma_law","chart","head_curve","euler"],"does":[[891.0678958333333,"probe is shown on the screen, written out."],[896.5588958333333,"probe is redrawn as the numbers it depends on change."],[896.5588958333333,"lam ticks to 140.0."],[899.5198958333333,"probe is redrawn as the numbers it depends on change."],[899.5198958333333,"lam ticks to 100.0."]]},{"start":902.2678958333333,"say":"Now draw in the yield strength. Two hundred and fifty megapascals for a common structural steel. And look at what the Euler curve does on the left of the picture. It climbs straight through it.","live":["lam_def","sigma_law","chart","head_curve","euler","probe"],"does":[[903.3708958333333,"yield_line is shown on the screen, written out."],[912.2988958333333,"probe is redrawn as the numbers it depends on change."],[912.2988958333333,"lam ticks to 70.0."]]},{"start":914.3268958333333,"say":"That part of the curve is a lie. It predicts a buckling stress well above the stress at which the material gives way. The column would have yielded long before it got there, and the derivation assumed elastic behaviour throughout.","live":["lam_def","sigma_law","chart","head_curve","euler","probe","yield_line"],"does":[[916.6718958333333,"point is shown on the screen, grown."],[919.1718958333333,"point is hidden from the screen."],[919.9228958333333,"yield_line is indicated — a transient flash."]]},{"start":927.8483958333333,"say":"The crossing point is where the two mechanisms trade places. Set pi squared E over lambda squared equal to sigma y and solve. Lambda critical is pi times the square root of E over sigma y, and for this steel that is about eighty nine.","live":null,"does":[[928.3588958333333,"cutoff is shown on the screen, written out."],[928.3588958333333,"crossing is shown on the screen, written out."],[937.5078958333334,"lam_c_eq is shown on the screen, written out."]]},{"start":945.0848958333333,"say":"So above about eighty nine, elastic buckling governs and the strength of the material barely matters. Below it, the column is stocky enough to reach yield first, and now the strength is what matters and the stiffness barely does.","live":["lam_def","sigma_law","lam_c_eq","chart","head_curve","euler","probe","yield_line","cutoff","crossing"],"does":[[947.0708958333332,"euler is emphasized."],[951.4238958333333,"euler is no longer emphasized."],[953.6878958333333,"yield_line is emphasized."],[957.1248958333333,"yield_line is no longer emphasized."],[958.6228958333334,"head_curve is hidden from the screen — left the board."],[958.6228958333334,"lam_c_eq is hidden from the screen — left the board."],[958.6228958333334,"lam_def is hidden from the screen — left the board."],[958.6228958333334,"sigma_law is hidden from the screen — left the board."]]},{"start":959.2228958333333,"say":"But that crossing is a sharp corner, and real columns do not have corners. Two things round it off, and both of them make the column weaker rather than stronger.","live":["chart","euler","probe","yield_line","cutoff","crossing"],"does":[[959.2228958333333,"head_real is shown on the screen, written out."],[960.4538958333333,"crossing is indicated — a transient flash."]]},{"start":970.0983958333333,"say":"First, every real column is slightly crooked to start with. So it begins bending from the very first increment of load, instead of waiting for a threshold and then choosing to bend.","live":["chart","euler","probe","yield_line","cutoff","crossing","head_real"],"does":[[970.0983958333333,"reasons is shown on the screen, written out."],[972.8728958333332,"reasons (the \"Every real column is slightly crooked, so it bends from the first increment of load.\" part) is emphasized."]]},{"start":982.3313958333333,"say":"Second, a rolled section cools unevenly and locks in residual stresses. Part of the cross section is already close to yield before you apply any load at all, so it softens early and the whole member follows.","live":["chart","euler","probe","yield_line","cutoff","crossing","reasons","head_real"],"does":[[986.7548958333333,"reasons (the \"Every real column is slightly crooked, so it bends from the first increment of load.\" part) is no longer emphasized."],[986.7548958333333,"reasons (the \"Rolled sections carry residual stresses, so part of the section yields early.\" part) is emphasized."]]},{"start":996.5038958333332,"say":"The consequence is this red curve. Real columns sit below both idealisations, and the gap is worst right around the crossing, in the intermediate range, which is where most practical columns actually live.","live":null,"does":[[996.5038958333332,"reasons (the \"Rolled sections carry residual stresses, so part of the section yields early.\" part) is no longer emphasized."],[998.1408958333333,"real is shown on the screen, drawn."],[1003.6438958333333,"point_2 is shown on the screen, grown."],[1006.1438958333333,"point_2 is hidden from the screen."]]},{"start":1010.5713958333333,"say":"So codes do not use either straight idealisation. They use a fitted curve of this shape, calibrated against tests, with Euler as the upper bound it approaches once the column is slender enough.","live":["chart","euler","probe","yield_line","cutoff","crossing","reasons","head_real","real"],"does":[[1015.2148958333332,"reasons (the \"Codes fit a smooth curve to test data, below both idealisations.\" part) is emphasized."],[1019.1158958333333,"euler is indicated — a transient flash."],[1019.8238958333333,"reasons (the \"Codes fit a smooth curve to test data, below both idealisations.\" part) is no longer emphasized."]]},{"start":1023.4663958333333,"say":"So, to gather it up. A slender column fails because the straight shape stops being the only equilibrium available, at a load of pi squared E I over the effective length squared.","live":null,"does":[[1026.6128958333334,"real is indicated — a transient flash."]]},{"start":1036.7438958333332,"say":"The end conditions enter only through where the points of zero moment sit. And the slenderness ratio tells you which failure you are actually designing against, with Euler an upper bound that gets more honest the more slender the column is.","live":null,"does":[[1041.6548958333333,"crossing is indicated — a transient flash."],[1046.7288958333334,"euler is indicated — a transient flash."],[1050.7456458333334,"chart is hidden from the screen — left the board."],[1050.7456458333334,"euler is hidden from the screen — chart left the board."],[1050.7456458333334,"probe is hidden from the screen — chart left the board."],[1050.7456458333334,"yield_line is hidden from the screen — chart left the board."],[1050.7456458333334,"cutoff is hidden from the screen — chart left the board."],[1050.7456458333334,"crossing is hidden from the screen — chart left the board."],[1050.7456458333334,"real is hidden from the screen — chart left the board."],[1050.7456458333334,"head_real is hidden from the screen — left the board."],[1050.7456458333334,"reasons is hidden from the screen — left the board."]]}]}]},"durationSeconds":1052,"chapters":[{"title":"Two Ways for a Column to Fail","startSeconds":0,"narration":"A steel strut one metre long will carry an enormous load. Make the same strut three metres long, from the same steel, with the same cross section, and it folds up under a small fraction of that. Nothing about the material changed. Here are two columns cut from one bar. The short one fails the way you expect. Push hard enough, the stress reaches the yield strength of the steel, and the material gives way. Now the long one. Push on it, and long before the stress anywhere in it comes near yield, it does this. It bows sideways. The material is still perfectly elastic. Unload it and it springs straight again. But as a structural member it has failed, because it will not take any more load than that. And the gap between the two is not small. Triple the length, and the load the column can take falls by a factor of nine. Nothing in the material noticed. This is buckling, and it is a failure of stability rather than a failure of strength. The question is what load it happens at, and why the length matters so much when the material does not. Here is the case we will solve. A straight column of length L, pinned at both ends, so each end is free to rotate but held against sideways movement. It carries an axial load P through the centroid. The material is linearly elastic with Young's modulus E, and the cross section has a second moment of area I about the axis it bends about. Those are the only four quantities the answer can depend on. The answer we are heading for is this. The critical load is pi squared E I over L squared, and the shape the column takes at that load is a single smooth half sine wave, zero at both pins and largest at midheight. I want both of those to fall out of the mechanics rather than be handed to you. And notice already what that formula does not contain. There is no yield stress in it anywhere. The load at which a slender column buckles has nothing to do with how strong the material is. Only with how stiff it is, and with how the length and the cross section combine. So let the column bend, write equilibrium for the bent shape, and see what the mathematics demands."},{"title":"Equilibrium of the Bent Column","startSeconds":144.641875,"narration":"Suppose the column is not straight. Not because something pushed it out of line, but suppose it simply is, by some small amount, and ask whether that bent shape can stand in equilibrium. The dashed grey line is where the axis used to be. Take a height x above the bottom pin, and call the sideways movement of the axis at that station v of x. Now cut the column there, and keep the piece below the cut. At the bottom, the axial load P acts along its original line of action, because that pin has not moved sideways at all. And there is no sideways reaction at the pin either, because there is no sideways load anywhere on the column. So the base hands the free body exactly one force, P, at zero offset from the original axis. That is the whole free body. One axial force at the base, and the internal shear and moment at the cut. Nothing else is acting on the piece. Take moments about the cut section. The load P is acting a distance v of x away from that section, so it applies a moment of P times v of x. The internal bending moment has to balance it exactly. So M of x plus P v of x is zero, and the bending moment at any section is minus the axial load, times how far that section has strayed from the line of action. Watch what that means as the cut moves. Near the ends the deflection is small, so the moment is small. At midheight the deflection is largest, and so is the bending moment. So the bending moment along the column has the same shape as the deflection itself, scaled by P. That is unusual. In ordinary beam bending you are handed the loading and you go and find the moment. Here the moment depends on the answer. And there is the loop that makes buckling what it is. Deflection produces bending moment. Bending moment produces curvature. Curvature produces more deflection. The load P is the gain around that loop. Now bring in the bending relation you already have. For small deflections, E I times the second derivative of the deflection equals the bending moment at that section. That is the small deflection form, and it is the honest limit of what we are doing here. As long as the bow stays small compared with the length, the curvature is well approximated by v double prime. Substitute the moment we just found. E I v double prime equals minus P v. Move everything to one side, and there is the governing equation. Divide through by E I, and give the group P over E I a name. Call it k squared. Then the equation reads v double prime plus k squared v equals zero, and k carries units of one over length. That is a second order linear equation with constant coefficients, and you already know every function that satisfies it. Solving it is what turns this into a statement about the load."},{"title":"The Critical Load and the Mode Shape","startSeconds":341.00456249999996,"narration":"v double prime plus k squared v equals zero. Every solution of that is a combination of a sine and a cosine of k x, so write the general one down with two constants in it. Now the end conditions. The bottom is pinned, so it cannot move sideways there. v of zero is zero. Put x equal to zero into the general solution: the sine term vanishes on its own, and what is left is B. So B is zero, and that leaves v equals A sine k x. The top is pinned too, so v of L is zero as well. Put x equal to L in, and you get A times sine of k L equals zero. Look hard at that line, because everything is in it. A product of two things is zero, so at least one of them has to be zero, and there are exactly two ways for that to happen. The first way is that A is zero. That gives v identically zero, the perfectly straight column. And notice it is available at every value of P. The straight configuration is always an equilibrium, and it never stops being one. The second way is that the sine itself is zero. Sine of k L vanishes when k L is a whole number of pi. And that is a condition on the load, because k squared is P over E I. Square both sides, substitute for k squared, and P L squared over E I equals n squared pi squared. So P equals n squared pi squared E I over L squared. These are not loads the column happens to reach. They are the only loads at which anything other than the straight shape is possible at all. Below the smallest of them, A has to be zero. The smallest is n equal to one. Pi squared E I over L squared. That is the Euler critical load, and it came out of demanding that a bent shape can stand in equilibrium, not out of any strength calculation. And the shape that goes with it is A sine of pi x over L. One half of a sine wave. Zero at both pins, largest at midheight, and no point of inflection anywhere in between. Now the part that catches people out. A is still undetermined. The mathematics has fixed the shape and said nothing whatever about how big it is. Here is the same mode at three different amplitudes. Every one of them satisfies equilibrium exactly, and every one of them does it at exactly the same load. That is the signature of the thing we are looking at. The higher values of n are genuine solutions too. n equal to two is a full sine wave, with a stationary point at midheight, and it needs four times the load. n equal to three needs nine times. In a bare column they never happen, because the column has already gone at n equal to one. But brace it sideways at midheight, and you have forbidden the first mode. The second one is what you get instead. Put all of that on one picture. Load runs up the vertical axis. The amplitude of the bow runs across the horizontal one, so the whole vertical axis is the straight column. Load it up from nothing. Below the critical load there is exactly one equilibrium at every value of P, and it sits on that axis at zero amplitude. At the critical load, a second branch appears. Every amplitude on that horizontal line is an equilibrium, at the one load, and the straight solution has stopped being unique. That splitting is called a bifurcation, and it is exactly what we mean when we say the column has buckled. Nothing broke. No stress reached any limit. The column simply ran out of reasons to stay straight. Now, that whole derivation used one particular pair of end conditions. Change them and the numbers change. What is worth seeing is that they change in a way you can read straight off the deflected shape."},{"title":"Where the Bending Moment Vanishes","startSeconds":593.553375,"narration":"Everything in that derivation came from two facts about the ends. They could not move sideways, and they could not carry bending moment, because a pin cannot. Change either one and the answer changes. But look at what the pinned answer really is. The moment is P times the deflection, and the deflection is zero at both pins. So the half wave runs between two points where the bending moment is zero. So here is the statement that replaces the table. The critical load is pi squared E I divided by the square of the distance between adjacent points of zero bending moment. Call that distance the effective length. The reason is simple. Between two such points, the deflected curve is a half sine wave with no moment at either end. That is precisely the pin ended problem we already solved, sitting inside a longer column, and it does not care what happens beyond those two points. And zero bending moment means zero curvature, so those points are the inflection points of the deflected shape. You can find them by looking at where the curve changes the way it bends. Take a column built in at both ends, so neither end can rotate. Load it until it buckles, and it takes this shape: vertical where it meets each support, bulging one way in the middle, and curling back near each end. Find the inflection points. The curve bends one way near the middle and the other way close to each support. It changes over here, and again here, at a quarter of the length in from each end. So the distance between them is half the length of the column. That middle half is a half sine wave with zero moment at both ends, and it has no idea it is not a pinned column of length L over two. Put L over two in place of L. The critical load is four times what the pinned column would carry. The factor K is one half, and it is not a number from a table. It is where the curvature vanishes. Now the other extreme. Built in at the bottom, completely free at the top. This is a flagpole. It buckles into a quarter wave: vertical at the base, and leaning out at the top with no moment there at all. There is only one point of zero moment on the column, and it is the free end itself. So where is the other one? It is not on the column. Reflect the shape through the fixed base. The reflected curve meets the real one with the same slope, so the two together make one smooth curve, and it is exactly the pinned half sine wave, of total length two L. So the effective length is two L, the factor K is two, and the critical load is a quarter of the pinned value. A flagpole is four times weaker than the same member pinned at both ends. One case is left, and it is the only one that is not a round number. Pinned at one end and built in at the other. The pin gives zero moment, so one of the two points is the pin itself. The other one is the single inflection point up here. Where it sits comes out of a transcendental equation rather than a fraction, and it lands at about seven tenths of the length. So K is roughly zero point seven. Here they all are together. In every row, K is nothing more than the fraction of the column that lies between adjacent points of zero moment, and the critical load is pi squared E I over that length squared. One warning worth carrying. These are the idealised values. A real connection is never a perfect pin and never perfectly fixed, so design codes pull these numbers back toward one. But the geometry is the reason they are what they are."},{"title":"Slenderness, and Where Euler Stops","startSeconds":815.0678958333333,"narration":"We have a critical load. To compare columns of different sizes we want a critical stress, so divide P critical by the cross sectional area. The cross section now appears twice in that expression, once as I and once as A. Those two are not independent, and the length that ties them together has a name. Define r, the radius of gyration, as the square root of I over A. It is the distance from the bending axis at which you could put the whole area and get the same second moment. For this rectangle, it is here. Substitute I equals A r squared. The area cancels, top and bottom, and what is left is pi squared E over the square of K L divided by r. That group is the slenderness ratio. Effective length over radius of gyration. It is dimensionless, it is the only geometry the answer needs, and it is what engineers actually mean when they call a column slender. So carry those two results across, and plot the critical stress against slenderness. It is a hyperbola: double the slenderness, and the critical stress falls by a factor of four. At a slenderness of two hundred, this steel column buckles at about fifty megapascals. Bring it in to a hundred and forty, and it is around a hundred. At a hundred, about two hundred. Now draw in the yield strength. Two hundred and fifty megapascals for a common structural steel. And look at what the Euler curve does on the left of the picture. It climbs straight through it. That part of the curve is a lie. It predicts a buckling stress well above the stress at which the material gives way. The column would have yielded long before it got there, and the derivation assumed elastic behaviour throughout. The crossing point is where the two mechanisms trade places. Set pi squared E over lambda squared equal to sigma y and solve. Lambda critical is pi times the square root of E over sigma y, and for this steel that is about eighty nine. So above about eighty nine, elastic buckling governs and the strength of the material barely matters. Below it, the column is stocky enough to reach yield first, and now the strength is what matters and the stiffness barely does. But that crossing is a sharp corner, and real columns do not have corners. Two things round it off, and both of them make the column weaker rather than stronger. First, every real column is slightly crooked to start with. So it begins bending from the very first increment of load, instead of waiting for a threshold and then choosing to bend. Second, a rolled section cools unevenly and locks in residual stresses. Part of the cross section is already close to yield before you apply any load at all, so it softens early and the whole member follows. The consequence is this red curve. Real columns sit below both idealisations, and the gap is worst right around the crossing, in the intermediate range, which is where most practical columns actually live. So codes do not use either straight idealisation. They use a fitted curve of this shape, calibrated against tests, with Euler as the upper bound it approaches once the column is slender enough. So, to gather it up. A slender column fails because the straight shape stops being the only equilibrium available, at a load of pi squared E I over the effective length squared. The end conditions enter only through where the points of zero moment sit. And the slenderness ratio tells you which failure you are actually designing against, with Euler an upper bound that gets more honest the more slender the column is."}]}}