The check that cannot see the beams

One column is enough to find the sway took a fixed-base portal frame, solved it, and found its sway of 10.38 mm again from one column’s two end moments and nothing else. The unit load that finds a deflection may be carried on any structure that can carry it — any structure will carry the unit load — and on the left column cantilevered alone it puts moment in that column and nowhere else, so the integral of the real moments against the virtual ones runs over one member.

A single-bay portal has three redundants and a handful of releases worth drawing. A building frame of four bays and ten storeys has a hundred and twenty, and the essay ended by asking whether the trick survives the multiplication: whether the sway of a tall frame can still be carried down one column line that reaches it, whether it stays a useful check when that line’s real moments come from beams that are cracked or joints that are semi-rigid, and whether there is a release for each deflection a designer checks that keeps the hand integral to one member.

A frame of forty columns #

The frame is concrete: four bays of 8 m, ten storeys of 3.5 m, 35 m tall, with 600 mm square columns continuous up every line and 400 × 700 mm beams at every floor. Every beam carries 40 kN/m and every floor 40 kN of wind at its windward end, half that at the roof. Solved by stiffness with every member’s bending and axial flexibility, it sways 15.7 mm at the top of its windward column line.

Carry a unit horizontal load at the roof of the windward line down that line as a cantilever — every beam cut where it meets the line, the line fixed at its base. The unit load then puts a moment $m(z) = H - z$ in that line’s columns and nothing in any beam or any other column. The roof sway is

$$\Delta = \sum_{\text{storeys}} \int \frac{M\,m}{EI}\,dz = \sum \frac{h}{6EI}\left[M_b(2m_b + m_t) + M_t(m_b + 2m_t)\right],$$

because the real moment $M$ is linear up each storey: two end moments and two products a storey, forty numbers for the whole line.

The real moments are the frame’s, and they look nothing like the unit load’s. The unit load’s moment is a straight line from nothing at the roof to 35 kN·m per kN at the base. The real moments are a zigzag: the ground storey’s columns bend in single curvature from their fixed bases, and every column above them bends in double curvature, each storey’s moment passing through nothing near its middle — the familiar picture of a sway frame, whose columns are held at both ends by beams.

Every line gives its own top’s sway #

Every column line’s integral reproduces the sway at its own top, to the precision the frame was solved to, and the five lines’ tops sway by slightly different amounts — 15.7 mm at the windward line, 15.2 at the leeward — because the beams between them shorten under the wind’s push. The trick does not need the left column, or the windward one. It needs a column line and the real moments in it.

And it survives every change to the frame drawn here. With the beams cracked to 0.35 of their gross stiffness the frame sways 33.3 mm and the line’s integral gives 33.3 mm; with the beam-to-column joints made springs of 50 MN·m per radian it sways 61.3 mm and the integral gives 61.3; with each base a spring of 200 MN·m per radian, 17.3 and 17.3 — once the base’s rotation is given its own term. The identity is exact whatever the beams and joints are, because the virtual system has no moment in any beam and passes through no joint spring. The unit load goes straight down a continuous column line; the beams’ stiffness, cracked or not, and the joints’ flexibility, rigid or not, enter the check only through what they did to that line’s real moments.

Five lines, five answers, one frame #

The five lines’ answers differ, and the differences are themselves a check. Between the windward line’s 15.7 mm and the leeward line’s 15.2 lie the four roof beams, pushed together by the wind on their way to the leeward columns; the half millimetre is their axial shortening, summed. A checker who works two lines rather than one gets two sways whose difference must be what the floor beams’ axial forces make of their lengths — a second identity, about the beams this time, though about their axial stiffness rather than the bending stiffness the sway depends on.

The two lines also divide the work differently. The middle column line carries larger moments than the windward one, 189 kN·m at its base against 126, because it has beams framing into it on both sides; its integral is built from larger numbers that cancel more. Any line will do, and the line with the smallest moments is the one whose products are easiest to get right by hand.

Why each line gets its own top #

The reason a column line’s integral comes out at that line’s top, and not at some average of the roof, is the equilibrium of the virtual system. A unit load at the top of line $i$, carried by line $i$ alone as a cantilever from its base, is in equilibrium with a unit reaction and a moment $H$ at that base and with nothing else. The principle of virtual work then says that the virtual load times the real displacement where it acts — the sway at the top of line $i$ — equals the virtual moments times the real curvatures, summed over the members the virtual system loads. There are no others, so there is nothing to sum but the line.

The real frame need not resemble the virtual one in any way. Its beams carry load, its other columns carry load, its joints turn; none of that enters, because the virtual system puts no moment where those things happen. The real curvatures of line $i$’s columns carry the whole effect of everything else in the frame, since they are what everything else has done to that line. It is the reduction theorem applied as far as it goes: the released structure can be as small as a single continuous line of members from the load to the ground.

The check cannot see the beams #

That is the property that makes the one-line check cheap, and it is also its limit.

The frame’s sway depends on its beams’ stiffness by a factor of more than three across the range engineers argue about: 15.7 mm with gross beams, 33.3 at the cracked allowance of 0.35, 52.0 at 0.2. Every one of those analyses passes the column-line check exactly, because each analysis’s column moments are consistent with its own sway. The column moments do change — the windward base moment almost doubles, from 126 to 235 kN·m — but they change together with the sway, and the check compares the two.

So the check confirms an analysis’s arithmetic, not its stiffness assumptions. It catches a sway that was read from the wrong node, a load case that was applied with the wrong sign, a member whose moments were reported in the wrong units — anything that makes the reported sway and the reported column moments disagree. It cannot catch beams modelled at their gross stiffness that should have been cracked, because the error is in both numbers equally.

What it does catch is an analysis that is not one analysis. Take the column moments from a run with gross beams and the sway from a run with cracked ones — the kind of mix that happens when the sway is read from one model and the column design from another — and the check reports 15.7 mm against a sway of 33.3: it fails by a factor of two, and the failure is the information. A sway that disagrees with its own column moments has come from a different structure, and the one-line check is the cheapest way there is to find out.

To make a check that sees the beams, the virtual system has to put moment in them, and then the checker has to supply the beams’ stiffness — the analysis’s, if the check is to pass, or the checker’s own, if it is to test the analysis’s choice. The portal method’s released structure, with hinges at the columns’ and beams’ mid-lengths, does that: its virtual moments fill every member, and the check then multiplies beam moments by a beam flexibility someone has to choose. That is the point of the column line and its limit at once. The release that needs no judgement about the beams is the release that cannot report on them. That is the same lesson one support too many drew from a redundant beam in the other direction: an indeterminate structure’s internal forces are a consequence of the stiffness assumed, and any check that takes them from the analysis inherits the assumption.

The account depends on the release #

The single portal’s essay found that different releases give the same total and different accounts of where it came from. In a tall frame the account given by one column line is stranger still. The ground storey contributes 27.4 mm to a roof sway of 15.7, and every storey above it a negative amount, between −1.8 and −0.5 mm. The reason is the zigzag: the virtual moment is large near the base and the real moments of the upper storeys pass through nothing in each storey, so their products are small and of whichever sign the double curvature leaves after weighting by $H - z$, which here is negative.

None of that is a statement about where the frame is flexible. The drawing that is right except for a rotation found the same thing in a truss’s deflected shape: a construction can be exact and its intermediate pictures still be pictures of a reference, not of the structure. The sway of a moment frame comes mostly from its beams bending, as one deflection without solving everything would find with the unit load carried by the frame itself; the column-line account has moved all of it into the columns, because that is the only place its virtual system has moment. The total is the deflection; the shares are the release’s. With springy joints the account’s ground storey grows to 70.1 mm and the storeys above go further negative, which is the same total redistributed through a softer frame.

A storey’s drift is mostly in the floor below #

The second half of the question was whether each deflection a designer checks has a release that keeps the integral to one member. The deflection checked most is a storey’s drift, the difference between the sway at a floor and at the floor below. Carried on one storey of one column as a cantilever from the floor below, a pair of unit loads gives that storey’s column’s own bending, measured from its tangent at the floor below — but that is not the drift. The floor below has turned, and its turning tilts the whole storey above it.

The ground storey’s 1.75 mm of drift is all its own: its columns stand on fixed bases and do not start out tilted. Every storey above it drifts mostly by the rotation of the floor below. The second storey’s 2.68 mm is 2.85 mm of the first floor’s rotation carried up 3.5 m and −0.18 mm of the column’s own bending, which is slightly backwards because a column in double curvature, measured from the tangent at its foot, bends back towards its chord. At the top, 1.03 mm of rotation and −0.54 of bending make 0.49.

A floor’s rotation is set by the beams framing into it and the columns above and below together, so a storey’s drift is not in that storey’s columns. There is no release that keeps a storey’s drift to one member, except at the ground storey: the unit-load pair must be carried down to a fixed point, through every storey below, and those storeys’ column moments enter the integral. The hand check of a drift is the hand check of the roof sway truncated at that floor.

A base that turns needs a line of its own #

One thing the virtual system does pass through is the base. A base that rotates under its moment — a footing on soil, a column bolted to a pad — is a spring, and the cantilevered column line’s unit load puts its full moment, H, into it. The base’s rotation $M_{base}/k$ times H is then a term of its own in the integral, as the pinned base that is not pinned found the base’s rotation a term of its own in a portal’s stiffness.

It is not a small term. At 178 MN·m per radian — a stiff footing — the frame sways 17.4 mm, the base’s term is 20.6 and the columns’ bending −3.2; at 32 MN·m per radian, 19.7 mm against a base term of 51.7. The frame’s sway hardly changes with the base, because the beams and the other columns redistribute the moment, but the column line’s account of it moves almost entirely into the base. Left out, the base term turns the check’s answer negative.

The ground storey by hand #

The windward column’s ground storey runs from 126.0 kN·m at its base to 25.0 at its top. The unit load’s moment there runs from 35 kN·m per kN to 31.5, and the column’s stiffness is $3.24 \times 10^5\ \mathrm{kN\,m^2}$, so the storey contributes

$$\frac{3.5}{6 \times 3.24 \times 10^5}\left[126.0\,(2 \times 35 + 31.5) + 25.0\,(35 + 2 \times 31.5)\right] = 1.80 \times 10^{-6} \times 15{,}239 = 0.0274\ \text{m},$$

27.4 mm. The nine storeys above add $-11.8$ mm between them, and the line’s total is 15.7. Forty end moments and forty products: an afternoon by hand in 1960 and a minute with a calculator now, against an analysis of 165 degrees of freedom.

What the identity assumes #

The column line is continuous. The virtual load must pass down the line without crossing a release. A column spliced with a pin, or a line that stops at a transfer beam, puts the virtual system into a beam, and the beam’s stiffness enters the check after all.

The real moments are the analysis’s. The identity holds for any compatible set of real deformations, and the analysis supplies one; the check is exact against the analysis, and only as good as the analysis is against the building.

Bending only. The columns’ shear deformation would add a term $\int V v/GA$, with the unit load’s shear of 1 down the line, and their axial deformation none, because the cantilevered unit load puts no axial force in any column. A frame whose columns shorten under the overturning moment sways partly by that shortening, and the column-line account moves that sway into the columns’ bending too.

A plane frame. A building’s frames are tied together by its floors, and a corner column belongs to two of them. The check survives the third dimension unchanged in form: a unit load in one direction carried down one column as a cantilever bends it about one axis only, so only the real moment about that axis enters, and a column bent both ways by a skew wind contributes the component the virtual load can see. Torsion of the whole building, which turns its floors, reaches the check only through what it does to that column’s bending.

One load case. The check is linear; under a combination it is the sum of the cases’ checks, which is the same thing.

Still open: the frame that leans on its gravity columns #

Every column here is part of the moment frame. Most buildings have a few frames that resist the wind and many columns that carry only gravity, pinned at every floor, leaning on the frames. A leaning column has no moment and so no share in any column line’s integral, but its axial load, acting through the sway, adds a moment to the frames that the first-order analysis here did not contain — the load that makes itself worse. Whether a one-line check of a second-order analysis needs a term for the leaning columns’ axial loads times the virtual system’s rotations, and whether the check can then see what a first-order analysis misses, is what the gravity columns ask of the frame.