Every joint balanced, and the frame still leaning

A joint is balanced when the moments meeting at it add to nothing. That is the whole of what a distribution enforces, it is enforced at every joint at once, and on a continuous beam it is enough — the answer walks in after four cycles and the beam is solved.

A frame that can lean is not solved by it, and the reason is a count.

Count the unknowns. Each joint that can rotate has a rotation, so this frame has two. Each storey that can translate has a displacement, so this frame has one. Three unknowns.

Count the equations a distribution writes. One per joint: two. The method has one equation fewer than the structure has freedoms, and the missing one is not a refinement or a second-order effect. It is the statement that the columns, between them, have to carry the horizontal load across.

What a balanced table can still be describing #

The way to see the gap is to do the arithmetic wrongly on purpose and look at what comes out.

Hold the frame against translation — imagine a prop from the beam to something immovable — and distribute. Nothing about the procedure changes. The fixed-end moments are the beam’s own $wL^2/12$, the distribution factors are the usual proportions of $4EI/L$, and the joints balance in the usual four cycles.

That figure is the defect, drawn. Every internal check passes. The moments at each joint sum to zero to as many decimal places as anyone cares to carry, the fixed-end moments were right, the carry-overs were right, and the answer is wrong by tens of per cent.

The failure is an omission rather than an error, which is why it survives checking. Nothing in a moment distribution table has a column headed horizontal equilibrium, so there is nothing to get wrong and nothing to notice missing. The same shape of defect runs through this collection: a check that reads what is on the page cannot report what is not on it, and an index that lists only what it was told about has the same blind spot one field over.

The equation that belongs to the storey #

Write it down and it is one line.

Each column carries a horizontal shear, and that shear is fixed by its two end moments: for a member with nothing applied along it, $V = -(M_{\text{top}} + M_{\text{base}})/h$. Sum the two columns and the total has to equal the load applied across the storey. Nothing about rotations appears in it.

Read the top bar of that figure carefully, because it contains the second surprise in this essay and it is not the one about props.

The no-sway pass does not leave 60 kN outstanding. It leaves 37.43. The columns, in a table that has no idea a horizontal load exists, are already delivering 22.57 kN of horizontal force to the storey — and they are delivering it because of the vertical load.

A symmetric load on an asymmetric frame #

Take the 60 kN away entirely and distribute the gravity load alone.

The mechanism is worth stating slowly, because “sway is caused by horizontal load” is the sentence that lets a designer skip the second pass.

The beam’s fixed-end moment is the same at both ends. What the two corners do with it is not the same, because the distribution factor at each corner depends on the column meeting the beam there. At the left corner the 5 m column contributes $4EI/5$ and takes 47.4 per cent of the out-of-balance; at the right corner the 3.5 m column contributes $4EI/3.5$ and takes 56.3 per cent. The stiffer member attracts more of it, which is the rule that runs through the whole subject and is the same one that decides which of two walls carries a building’s wind.

So the two corners end up with different moments, the two columns end up with different shears, and the difference has nowhere to go but sideways.

This is not a curiosity about stepped bases. Any frame whose columns differ — in length, in section, in end condition — sways under gravity, and most real frames differ in at least one of those. A symmetric frame under a symmetric load is the one case where the effect vanishes, and it is the case every worked example is set on.

The numbers the two corners were given #

The unequal shares are worth writing out, because they are the whole of the gravity-sway mechanism and they are fixed before any arithmetic starts.

At the left corner two members meet: the 5 m column at $4EI_c/5 = 32{,}000$ and the beam at $4EI_b/9 = 35{,}556$, in the units the frame is drawn in. The column’s share is 0.474 and the beam’s 0.526. At the right corner the 3.5 m column offers $4EI_c/3.5 = 45{,}714$ against the same beam, and the column’s share is 0.563.

The two columns take 47.4 and 56.3 per cent of their corners’ out-of-balance, and the difference between those two numbers is the reason the frame moves. Nothing else is asymmetric about the load: the beam’s fixed-end moment is $10 \times 9^2/12 = 67.5$ kNm at each end, applied in opposite senses, exactly as it would be on a symmetric frame.

That is a general result about distribution factors and it goes the other way too. Make the two columns the same and the shares are the same, the shears cancel, and the gravity sway is exactly zero — which is why every textbook portal has equal legs and why the effect is easy to have never met.

It also says which frames to be careful of, and the list is longer than “stepped bases”. Two columns of different section, because one carries a crane rail. A frame with a mezzanine tying one column and not the other. A bay at the end of a building where one column is stiffened by a gable and the other is not. Each is a frame whose two sides distribute differently, and each sways under its own dead load before anybody has thought about wind.

An arbitrary displacement, and what it is worth #

The second pass starts from the wrong end. It has no load in it, no fixed-end moment from anything applied, and no idea how big it should be.

What it has is a displacement. Push the top of the frame sideways by some amount, hold both joints against rotation while doing it, and each column acquires end moments of $6EI\Delta/h^2$ — the moments needed to keep a member’s ends parallel while its chord rotates. The beam, whose ends move together, acquires none.

The $1/h^2$ is the whole of why a short column is the dangerous one, and it arrives here rather than being asserted. Halve a column’s height and it attracts four times the sway moment, having been drawn as the smaller and cheaper member. The same relation reappears wherever a stiff element is put beside a flexible one and expected to share: a short spandrel between two long ones is the same arithmetic in a different plane.

Now the division. The sway pass, at the size drawn, supplies −102.80 kN of storey shear. The prop was carrying 37.43. So the pass is scaled by

$$\lambda = -\frac{37.43}{-102.80} = 0.3641$$

and added to the first.

Nothing about λ was available before the first pass was finished. It is a ratio of two quantities, both of which are outputs, and it is the reason the procedure cannot be run as a single table however many cycles are spent on it.

What the correction actually changes #

It is tempting to think of the second pass as a correction in the ordinary sense — a modest adjustment to an answer that was nearly right. On this frame it is not.

The tall column’s base moment goes from 49.62 kNm to 18.28: a change of 63 per cent, and a change of proportion rather than of size, since the two columns’ shares are completely rearranged. The right corner goes from 122.14 to 157.81. The left corner goes from 99.24 down to 72.96. One end of the frame gets worse by 29 per cent and the other gets better by 26 per cent, which means a designer who stopped after the first pass would have sized one column generously and the other one short.

There is no rule of thumb that recovers this. The scaling factor depends on the ratio of two storey shears, one of which is an artefact of holding the frame and the other of which came from a displacement chosen for convenience, and neither has a typical value.

That curve is the design consequence, and it inverts an intuition. The column that is shorter is stiffer, the stiffer member attracts the load, and the member a drawing makes look minor is the one the storey pushes hardest. It is the same statement as a short brace attracting a frame’s shear and the same statement as a stubby wall taking a building’s torsion, and the arithmetic here is where it can be watched happening.

The frame that carries both loads and does not lean #

The dip is the interesting part of that plot, and it is not the curve running out of resolution.

The hold force is the difference between what the storey needs and what the vertical load has already supplied. Those two contributions act in opposite senses on this frame: the wind pushes it one way and the unequal columns push it the other. As the second column is shortened the gravity contribution grows, and somewhere it grows to exactly 60 kN.

A frame with no symmetry, under two loads with nothing in common, that does not sway. It is worth being clear about what is and is not remarkable in that.

It is not a design technique. Nobody proportions a portal so that its gravity sway cancels its wind sway, because the wind is not one number, it blows both ways, and the cancellation holds for exactly one load case out of the hundreds a frame is checked for. Reverse the wind and the two contributions add instead: the hold force at that ratio becomes 120 kN, which is twice what a symmetric frame would have shown under the same load.

What it is, is a demonstration that the sway correction is not a wind correction. It is a correction for one equation being missing, the equation has two sources on this frame, and either of them alone would have been enough to need it. A designer who reaches for the second pass when there is horizontal load and skips it when there is not has understood the procedure as a rule about wind, and this ratio is where that reading gives an answer that is exactly right for exactly the wrong reason.

It is also the cleanest available answer to the question of what λ means. It is not a correction factor with a typical size, it is not bounded, and it is not positive: at ratios below 0.476 it comes out negative, and the second pass is subtracted. A quantity that changes sign has no rule of thumb, which is the same thing an influence line says about where to stand a load and the reason both have to be computed rather than assumed.

The base condition moves the whole picture #

Every number above is for fixed bases. Pin them and the frame is a different structure.

With pinned bases the columns lose their base moments entirely, so the storey shear has to be carried on the top moments alone, and the frame is far softer: the sway is 20.0 mm against 3.79. The corner moments become 12.7 and 218.89 kNm, which is a redistribution so complete that the two ends of the same beam are now different members.

Three things in that figure are worth separating, because they are usually collapsed into one sentence about pinned bases being softer.

The frame did not get weaker, it got less able to share. A fixed base is a second place a column can resist rotation, and taking it away leaves the beam as the only restraint at the top and nothing at all at the bottom. What the columns lose is not strength but a route, which is the same observation an effective length calculation makes about the same four end conditions and reaches by a completely different argument.

The asymmetry got worse rather than better. With fixed bases the two corner moments differ by a factor of 2.16; with pinned ones by 17. Removing restraint did not average anything out. It concentrated the frame’s response into the one path still available, and the shorter column — being the stiffer of the two remaining paths — took almost all of it.

And the sway grew by more than the moments did. Displacement is the quantity that is sensitive to a base condition, which matters because the drift limit is usually the check that governs a portal and the moment check usually is not. A frame designed on its moments and detailed with a base condition it did not have can fail the check nobody was watching.

The practical version of that is a caution rather than a calculation. A base detailed as pinned and built as something stiffer, or the reverse, does not change this frame’s answer by a few per cent — it changes which member the frame fails at. A base plate is never as pinned as it is drawn, and this is one of the places where the difference is not a refinement.

Both passes have to be finished, and one of them decides the other #

The two passes converge independently, and it is easy to assume that stopping each of them early costs each of them a little accuracy.

It costs more than that, because λ is computed from them.

Two cycles is not the engineering answer here that it is on a beam. The error compounds: a hold force measured from an unconverged first pass is scaled by a factor measured from an unconverged second pass, and the product moves further than either.

The comparison in that figure is worth naming for what it is. The curve is the distance from an answer obtained by writing three simultaneous equations in two rotations and a displacement and eliminating — slope-deflection, which shares no arithmetic with any of the above. The two agree to about $4 \times 10^{-13}$ kNm once both have converged. That agreement is the only evidence in this essay that the sign of the storey-shear equation is right, and it is the equation the whole procedure exists to satisfy.

The approximation that assumes this away #

There is a hand method one step cruder than this one, and the comparison says exactly what the second pass is buying.

The portal method makes a sway frame determinate by asserting a point of contraflexure at the mid-height of every column and the mid-span of every beam. That assertion supplies the missing equations directly — no hold force, no second pass, no scaling factor — and the analysis becomes statics.

What it costs is visible on this frame. A point of contraflexure at mid-height means a column’s two end moments are equal and opposite, so each column’s share of the storey shear is decided by nothing but the assumption. On the portal above the tall column’s end moments are 18.28 and 72.96 — a ratio of four, not one — and the contraflexure sits 1.0 m above the base rather than at 2.5 m. The approximation is not slightly out; it has put the point of zero moment in the wrong fifth of the member.

The two methods are answering the same question and differ in what they are willing to assume. The portal method assumes the shape of the answer and gets statics in return. Moment distribution assumes nothing about the shape and pays for it with a quantity — the hold force — that has to be measured from a completed table and then removed. Neither is a refinement of the other, and a frame checked by both is being told something useful by the disagreement, which on this one is 63 per cent at a column base.

One storey has one unknown, and ten have ten #

The frame does not shorten. Every column here is treated as rigid along its own axis, so the beam translates without dropping. A real frame’s columns shorten under load by different amounts, the beam tilts, and the differential shortening is a separate calculation this one cannot see.

The sway is small and the geometry is the undeformed one. Every moment above is computed on the frame as drawn. At 3.79 mm over 5 m that is fine; at the drifts a tall frame actually reaches, the vertical load acting through the displacement adds moment of its own, and that second-order term is a separate amplification applied on top of everything here.

One storey has one unknown; ten storeys have ten. A multi-storey frame needs one sway pass per storey, and the passes interact — a sway at the third floor changes the storey shear at the second. The hand procedure then becomes a small simultaneous solve over the sway passes, which is the point at which the method’s advantage over the matrix it replaced disappears entirely, and is the honest reason the technique is now a teaching object.

The bases do not move. A fixed base on a real foundation rotates, and a rotation there feeds straight into the storey equation. The condition is not fixed or pinned but a spring, and the frame’s answer sits somewhere on the line between the two sets of numbers above.

The 3.79 mm is on none of these figures #

The moment diagrams above are drawn on the frame’s members, which makes them readable and hides the thing that actually happened: the frame moved. The 3.79 mm is not on any of these figures, because at the scale that makes the members visible it is a fifth of a line width — and it is the quantity the entire second pass was about.

The other absence is the prop. It appears in the captions as a number and never as an object, because it is not an object: it is a term in an equation that was written down to be removed. A reader who wants to see it has to picture a frame with a hand on it, and the whole of the correction is the arithmetic of taking the hand away.

The assumption underneath the arithmetic #

Everything here rests on superposition — that the moments from the held frame and the moments from the swayed frame can be added, and that the sum is the moments of the frame that was neither held nor swayed.

That is true for a linear elastic structure and it is exactly as true as that condition is. A frame with a yielded member is not one, and once a plastic hinge has formed the shares are decided by strength rather than stiffness — at which point the two-pass procedure is describing a structure that no longer exists. The sway correction is an elastic technique, and its answer is the one the frame gives before anything gives way.

The correction Cross’s paper does not contain #

Later rungs on this anchor: the multi-storey sway procedure and the simultaneous solve over storeys that it turns into. Frames with inclined members, where the sway displacement of one member is not the sway displacement of another and the geometry of the mechanism has to be worked out before any distribution starts. Haunched members, where both the stiffness and the carry-over are properties of a taper. And the version of all of this with the vertical load acting through the drift, which is where the elastic answer above becomes the first term of an amplification.

Hardy Cross’s paper of 1930 does not contain the sway correction. It is a beam method as published, ten pages long, and the extension to frames free to translate was worked out over the following decade by other hands — which is the usual shape of a great method’s history and worth knowing about this one in particular. The idea that made it great was that the arithmetic does not grow with the structure’s connectivity: a joint with four members costs no more than a joint with two. Sway is the exception, and it is the exception precisely because it is not a property of a joint at all. It belongs to the storey, and there is no way to write it down one joint at a time.