One column is enough to find the sway

Any structure will carry the unit load found the freedom that makes the unit-load method more than a recipe. To find one deflection of an indeterminate structure, the real internal forces must be the real ones — found by solving the structure properly — but the virtual system, the forces a unit load at the deflection’s point would cause, need only be in equilibrium. It may stand on any statically determinate structure cut out of the real one, and the integral of real against virtual comes to the same deflection whichever is chosen. That essay worked in one straight member, a propped cantilever, where the choices were a cantilever, a simple span or a hinge.

A frame has more members and more kinds of internal force. A unit load on a portal frame bends its columns and its beam, pushes and pulls them axially, and shears them, and the integral grows a term for each:

$$\Delta = \sum_{\text{members}} \int \left( \frac{M\,\bar m}{EI} + \frac{N\,\bar n}{EA} + \frac{V\,\bar v}{G A_v} \right) ds.$$

The release then decides which of those terms appear at all. That is the subject here: what the choice changes, what it cannot change, and what follows about the habit of reading the integral term by term.

A portal, and the one number to reproduce #

Take a portal 12 m wide and 6 m high, fixed at both bases, with an IPE 500 beam on two HEB 300 columns: a 25 kN/m load on the beam and 40 kN of wind at the top of the left column. Solved by stiffness, with every member allowed to bend, stretch and shear, it sways: the top of the left column moves 10.38 mm to the right.

That number is the target. Every calculation below integrates these real internal forces — fixed by the frame as built — against the forces a unit horizontal load at the top of the left column would cause on some structure, and every one of them has to come back to 10.38.

Four structures to carry one unit load on #

The portal has three redundants, so it can be made determinate in many ways, and four are drawn. Remove the right base, and the left column is a cantilever with the beam and the right column hanging off it as a bracket: a unit load at the top of the left column goes straight down that column in bending and shear, and nothing else in the frame feels it. Remove the left base instead, and the load must travel along the beam as an axial force to the right column, which carries it down. Put pins at both bases and at midspan, and the three-hinged frame shares the load between both columns, with an overturning couple carried as axial force up one column and down the other. Put a pin at the left base and a roller at the right, and the left column takes the whole horizontal reaction while the beam carries the overturning moment across the frame.

The reduction theorem says all four are admissible. The figure shows how differently they are shaped: in one the virtual moment is confined to a single member, in another it fills all four.

The totals agree and the shares do not #

Integrate, member by member, and the five totals are identical to the micrometre: 10.38 mm each. Nothing else about them is.

Carried on the left column alone, the whole of the sway is the left column’s — the beam and the right column carry no virtual force, so their integrals are zero however hard the real frame works them. Carried on the right column alone, nearly all of it is the right column’s, with a sixth of a millimetre from the beam’s axial stretch. On three hinges the left column contributes −15.78 mm, a negative amount, and the right column 19.84. On the pin and roller the left column contributes −31.43 mm and the beam 41.65.

The unit load carried on the frame itself — the real, indeterminate portal, which is also an admissible structure — gives a fifth account: −3.72 from the left column, 11.28 from the right, 2.82 from the beam.

So a member’s “contribution to the sway” ranges from −31.4 mm to +10.4 mm for the left column alone, depending on nothing but a bookkeeping choice made after the frame was solved. This is the frame version of the two products that share nothing but their area: the integral is a pairing of two moment diagrams, one fixed and one chosen, and the pairing is free to put the answer anywhere the chosen diagram has something to multiply.

The whole sway from one column #

The left-column release is worth having for its own sake, because it turns a frame calculation into a cantilever one. The virtual moment in the left column is the unit load times the distance above the section, $\bar m = -(h - s)$, the virtual shear is one, and the rest of the frame is absent. The real moment in the left column varies linearly between its end values, and the real shear is its slope. So

$$\Delta = \int_0^h \frac{M(s)\,(h - s)}{EI}\,ds + \frac{V\,h}{G A_v},$$

which needs exactly three numbers from the real analysis: the left column’s base moment, its top moment and its shear.

From the frame those are 26.9 kN·m at the base, 149.2 kN·m (of the opposite sign) under the beam, and 29.3 kN. The integral of a linear moment against a triangle is a sixth of the length times the usual combination: with $M(s) = 26.9 - 29.35\,s$ and $\int_0^6 (6 - s)\,ds = 18$, $\int_0^6 s(6 - s)\,ds = 36$, the bending term is $(29.35 \times 36 - 26.9 \times 18)/EI = 573/52{,}857 = 10.84$ mm, with $EI = 52{,}857\ \text{kN·m}^2$ for the column. The shear term is $-29.35 \times 6/(G A_v) = -176/380{,}700 = -0.46$ mm, the shear in this column acting against the unit load’s. Together, 10.38 mm.

That is a check of a computer’s sway that needs no frame analysis at all — only three numbers it already printed for one member, and a cantilever formula. It does not check the three numbers; it checks that the sway the program reports is consistent with the forces it reports, which is the check worth making by hand on any output.

How much of the sway is axial #

The same freedom applies to the three kinds of term. A designer asks how much of a frame’s sway comes from the columns’ axial shortening — whether it is safe to leave out, whether a stiffer column section would help. Read off the integral, the answer depends on the structure the unit load was carried on.

On the left column alone, no virtual axial force exists anywhere, so the axial term is zero and the shear term is negative, −0.46 mm. On the right column alone the axial term is 0.34 mm and the shear term 1.09. On three hinges, 0.19 and 0.42; on the pin and roller, 0.02 and −0.36. The bending term absorbs the difference each time and the total does not move. A negative shear term is not shear deformation making the frame stiffer. It is a product of a real shear and a virtual shear of opposite sign, and it means nothing on its own.

This answers the question any structure will carry the unit load left open, which was whether a release can be chosen so that the virtual system’s axial forces vanish in the members whose axial stiffness is least certain. For the sway it can: the left column alone carries the unit load down by bending and shear only, and the integral then contains no $1/EA$ anywhere. That is genuinely useful when a member’s axial stiffness is uncertain — a column with a splice, a beam with a slotted end — because the deflection can then be computed without that stiffness appearing in the integral. What it cannot do is remove that stiffness from the problem. It is still inside the real moments, which came from an analysis that needed it.

The only terms that are slopes #

There is one structure whose terms do mean something, and it is the frame itself. Re-solve the portal with every member’s axial flexibility multiplied by a factor, and the sway rises along a gently curving line; at the frame as built its slope is 0.18 mm per unit of the factor. That is exactly the axial term the unit load on the frame itself gives, and the shear term, 0.36 mm, is the slope against the shear flexibility in the same way. The other structures’ terms are not slopes: the right column alone would predict 0.34, the left column alone zero, and neither is how the sway actually responds.

The reason is the one Castigliano’s theorem states. The sway is a derivative of the frame’s complementary energy, and when a member’s flexibility changes, the real forces redistribute as well — but because the real forces already make the energy stationary, their redistribution has no first-order effect, and the change in the sway is the real force times the frame’s own unit-load force, integrated over that member. That second factor has to be the force the unit load produces in this frame. A release’s virtual force is a different force, and multiplying by it gives the right total only because the release’s errors cancel across members.

So reading a deflection term by term is legitimate only when the virtual system is the structure’s own response to the unit load. In a determinate structure there is no choice — the only admissible virtual system is the structure’s own — which is why reading a truss deflection member by member is safe and finds members that are not worth stiffening. In an indeterminate one, the release that makes the integral cheapest is the one that makes its terms meaningless.

What the frame’s own system costs #

The meaningful terms have a price, and it is the whole analysis again. The unit load on the frame itself is a load case on an indeterminate structure: to know its virtual moments, axial forces and shears in every member, the portal has to be solved a second time, with the unit load in place of the wind and the gravity. For a computer that is free — the stiffness matrix is already factorised, and a second load case costs a back-substitution. By hand it is the reason the method was invented the way it was. The point of the reduction theorem was that the second solution can be skipped, and the left column alone skips it completely.

So there are two uses of the integral and they want opposite structures. To find a deflection, or check one, carry the unit load on the simplest structure that can carry it, accept that its terms are bookkeeping, and read only the total. To explain a deflection — to say which member to stiffen, or whether axial shortening can be ignored — carry it on the frame itself and pay for the second analysis, because only then is each term the rate at which the deflection responds to that member’s flexibility. Mixing the two, by reading the terms of a cheap release as if they were sensitivities, is the mistake the members figure was drawn to prevent.

There is a third account, independent of any unit load, for anyone who wants to divide the frame’s deformation among its members without choosing a structure. The strain energy each member stores under the real loads, $\tfrac12\int (M^2/EI + N^2/EA + V^2/GA_v)\,ds$, is fixed by the real forces alone. Here it is 3.19 kJ in all: 12 per cent in the left column, 57 in the beam and 31 in the right column. But it is an account of the work done by all the loads — the wind and the 300 kN of gravity together — and not of the sway, which is the wind’s own displacement. The wind’s work, half of 40 kN times 10.38 mm, is 0.21 kJ, a fifteenth of the energy stored. A member’s energy share answers “where is the frame strained”, and the sway’s sensitivities answer “what would make this deflection smaller”; they are different questions, and neither is answered by a release.

Down the middle, a column is unavoidable #

The freedom has limits, and the midspan deflection shows where. A unit load acting down at the middle of the beam must reach the ground, and only the columns reach it. Every admissible structure therefore carries it down one column or both as an axial force, and every one of them has a column axial term: 0.27 mm on the left column alone, 0.30 on the right, 0.46 on three hinges. No release removes it.

That is the general answer to “can it always be done”, and it is a question about the structure’s topology rather than its numbers. A virtual system can avoid a kind of force in a member only if some path from the load to the supports avoids needing it. A horizontal load at the top of a column has such a path — down the column in shear and bending — and a vertical load on a beam does not, because the only members that touch the ground are columns standing in the load’s direction. The midspan figure makes the same point as the sway figure, differently: on the left column alone the left column contributes 41.9 mm of a 32.3 mm deflection and the beam takes back 9.6, and nobody would mistake those for the members’ shares of anything.

Linear, small, and with the right real forces #

The frame is linear and its deflections small. The unit-load integral rests on superposition, and the reduction theorem on the real forces being in equilibrium on the undeformed shape. A slender frame whose sway is amplified by its own axial loads — the second-order effect — has real forces that depend on the deflection, and the integral then needs those forces, not the first-order ones.

The shear areas are a convention. The shear term uses $G A_v$, with $A_v$ the web’s area for an I-section. That is the right stiffness for a long member and an approximation near joints, where the web’s shear field is not uniform. The term is small here — 0.36 mm of 10.38 — and its uncertainty is smaller.

And the real forces must be right. Every release reproduces the deflection only because the real forces satisfy compatibility in the actual frame. Feed the integral real forces from a wrong analysis — a missed load, a joint modelled as rigid that is not — and each release returns a different number. That is a feature: two releases that disagree are a proof that the real forces are not the frame’s own.

Joints, foundations and P-delta #

They cannot show the joints. Every joint here is rigid, and a real beam-to-column connection is neither pinned nor rigid; its rotation adds a term to the integral at each joint, and that term, too, divides differently between releases.

They cannot show the foundations. A fixed base is an idealisation, and a base that rotates under its moment adds a spring term which, on the left column alone, multiplies the base moment by the unit load’s base moment, 6 kN·m per kN.

And they cannot show the second-order sway. Under its gravity load the frame’s columns carry 142 and 158 kN, and the sway amplifies itself a little; how much is a different calculation, and one whose answer is, unlike any of these terms, independent of how it is booked.

Same total, different accounts #

The unit load may stand on any structure that can carry it, and in a fixed portal five of them give the sway exactly: 10.38 mm.

The members’ shares are not properties of the frame. The left column contributes −3.72 mm on the frame itself, all 10.38 when it stands alone, −15.78 on three hinges and −31.43 on a pin and roller.

One column is enough. Its base and top moments and its shear give the whole sway by a cantilever formula: 10.84 mm of bending less 0.46 of shear.

Only the frame’s own terms are slopes. Its axial term, 0.18 mm, is how fast the sway grows with the axial flexibility; no release’s is. And a vertical load cannot be carried without an axial force in a column, so not every term can be released away.

Still open: the frame with more bays than releases worth drawing #

A single-bay portal has three redundants and a handful of sensible releases, each of which can be drawn. A building frame of four bays and ten storeys has a hundred and twenty, and the number of determinate structures inside it is astronomically larger. The left-column trick generalises — the sway of any storey can be carried down the one column line that reaches it, in bending and shear — but whether it stays a useful check when that column line’s real moments come from a frame whose beams are cracked, or whose joints are semi-rigid, and whether there is a release for each kind of deflection a building’s designer actually checks that keeps the hand integral to one member, is the question the multi-bay frame puts to this one.