The load that is not a load

Every application of the column analogy so far has begun with a load. The frame is released, the released structure carries the load and produces a moment diagram $M_s$, and that diagram is loaded onto the analogous column as a distributed load whose three resultants are the section’s direct stress and two bending stresses. Take the load away and there is no $M_s$, so there is nothing to load the section with, and the method appears to have nothing to say.

It has something to say about a case with no load in it at all, and the case is common: one foot of a frame goes down. A fixed-base portal is three times redundant, so it resists the movement, and moments appear throughout a structure carrying nothing.

The section is loaded by the displacement #

The three compatibility conditions have not changed and their right-hand sides have. Cut the frame at mid-span, apply the three redundants on a rigid arm at the elastic centre, and the conditions are still about what the cut does — except that the cut is now required to open by exactly the amount the settlement opens it:

$$A\,M_0 = -EI\,\Delta\theta, \qquad I_x H + I_{xy} V = -EI\,\Delta u_x, \qquad I_{xy} H + I_y V = -EI\,\Delta u_y .$$

The left-hand sides are the same three expressions. The right-hand sides used to be integrals of $M_s$ over the section; they are now the three components of the relative displacement the settlement imposes on the released structure, multiplied by $EI$.

That is what “a load on the analogous column” means here. In the load case the section is loaded by a distributed pressure of intensity $M_s$; in the settlement case it is loaded by whatever combination of direct force and couple produces the right $P$, $M_x$ and $M_y$ — and for a settlement those are not an integral of anything, they are three numbers read off a rigid-body displacement.

For a foot that goes straight down, the released structure’s right half translates down by $\Delta$ and does not rotate. So $\Delta\theta = 0$, $\Delta u_x = 0$, $\Delta u_y = -\Delta$, and

$$M_0 = 0, \qquad \begin{bmatrix} I_x & I_{xy} \\ I_{xy} & I_y \end{bmatrix}\begin{bmatrix} H \\ V \end{bmatrix} = EI\begin{bmatrix} 0 \\ \Delta \end{bmatrix}.$$

The direct stress is exactly zero. Not small — zero, because a pure translation imposes no relative rotation and the direct stress is conjugate to rotation. On a symmetric frame $I_{xy}$ is zero as well, so $H$ vanishes too and the whole answer is one bending stress, $V = EI\,\Delta/I_y$, giving a moment $V(x - \bar x)$ that is linear in $x$ and antisymmetric about the frame’s centre. That is the diagram in the figure: equal and opposite at the two feet, zero at the crown.

A settlement case is therefore simpler in the analogy than a load case, which is the opposite of what it is by any other method.

The check, which is a different calculation entirely #

The two routes have almost nothing in common. The analogy computes an area, a centroid and three second moments of a fictitious section, then solves a two-by-two whose right-hand side is a displacement. The flexibility check assembles a stiffness matrix for twenty-two members, removes one restraint, applies a unit force, inverts, reads one number out of the displacement vector, and scales. Neither borrows a step from the other, and they agree to the precision of the arithmetic.

The unsymmetric case is the one worth checking, because it is where the two coupled equations both do work. With columns of 5 and 3 metres the product of inertia is not zero, so $H$ is not zero either: a foot that moves straight down develops a horizontal thrust as well as a vertical force, and the moment diagram is no longer antisymmetric — −6.74 at the left foot against +3.43 at the right.

The scaling that runs the wrong way #

Two lines, and between them is the whole difference between a load and an imposed displacement.

The load’s line is flat, and the reason is worth stating carefully because it is easy to half-remember. Stiffening every member of a frame by the same factor changes no ratio of stiffnesses, and a load is distributed among redundant paths by ratios. So the moments are unchanged: a frame twice as stiff under the same load carries the same moments and deflects half as far. Stiffening one member does change the ratios and does move the moments — toward the member that was stiffened — but that is a different operation.

The settlement’s line goes through the origin. The settlement imposes a curvature on the frame, and a moment is $EI$ times a curvature. Double the rigidity and the same geometry of deformation costs twice the moment. A frame of zero stiffness would follow its foot down and carry nothing.

So the two cases respond to the designer’s most common instinct in opposite directions. A member that is failing under load is made bigger; a member that is failing under a settlement is made bigger and fails harder. The lines cross, on this frame, at about $EI = 205{,}000\ \mathrm{kN}{\cdot}\mathrm{m}^2$, above which the settlement is the larger of the two cases.

That is not a reason to build flimsy frames, and stating it as one would be a caricature. It is a reason to know which of the two a member is sized by, because the two cases point in opposite directions and only one of them is on the load schedule.

What ten millimetres is worth #

Ten millimetres is not a failure of a foundation. It is within the movement a spread footing on a firm soil is expected to make as the load comes on, and it is a fraction of what the same footing will do over decades. Thirty millimetres is a bad day and not a remarkable one.

Nine and a half per cent is also not a failure. It is the size of the effect that matters here rather than its sign: it is comparable with the difference between one load combination and the next, it appears nowhere on the load schedule, and it is not covered by any factor applied to the loads. A partial factor of 1.35 on a permanent action multiplies the load case by 1.35 and multiplies the settlement case by exactly one.

There is a second reason the number is easy to lose, and it is the one the essay on differential settlement is about: only the difference between two supports does anything. A frame whose two feet both go down twenty millimetres develops no moment at all, because a rigid-body translation is not a deformation. Ground investigation reports settlement; structures respond to differential settlement; and the difference between the two is a number nobody measures.

The whole of it, once, by hand #

For the symmetric portal, $L = 8$ m, $h = 5$ m, all members at $EI = 20{,}000\ \mathrm{kN}{\cdot}\mathrm{m}^2$, right foot down $\Delta = 10$ mm:

The analogous column is the centreline at unit width, so $A = 5 + 8 + 5 = 18$ m and its centroid is at $\bar x = 4$ m, $\bar y = 65/18 = 3.611$ m — which does not enter the answer, because $H$ is zero and only the vertical axis is used. About the vertical axis through the centroid,

$$I_y = \int (x - \bar x)^2 ds = 2 \times 5 \times 4^2 + \frac{8^3}{12} = 160 + 42.67 = 202.67\ \text{m}^3 .$$

With $\Delta\theta = 0$ and $\Delta u_x = 0$, the first two redundants vanish and the third is

$$V = \frac{EI\,\Delta}{I_y} = \frac{20{,}000 \times 0.010}{202.67} = 0.987\ \text{kN},$$

which is the vertical force the settlement develops, and the moment anywhere is $V(x - \bar x)$:

$$M(\text{left foot}) = 0.987 \times (0 - 4) = -3.95\ \text{kN·m}, \qquad M(\text{crown}) = 0.987 \times 0 = 0 .$$

Three lines of arithmetic for a case that by the force method is a three-by-three with a displacement on the right-hand side. The whole saving is that $I_y$ is a property of the frame and not of the load, so it is computed once and used for the load case, the settlement case and anything else the frame is asked.

And the units are worth watching, because this is the one case in the analogy where they bite. In the load case $EI$ cancels out of the answer if every member has the same value, so it is usually carried as one and forgotten. Here it does not cancel — it multiplies — so a frame analysed in units of $EI = 1$ gives a settlement moment of $0.010/202.67 = 4.93 \times 10^{-5}$, which has to be multiplied by 20,000 before it means anything.

The unsymmetric frame, where both redundants wake up #

A foot that moves straight down, on a frame that is not symmetric, produces a horizontal reaction. That is the product of inertia doing its work again: the second and third conditions are coupled, so a displacement imposed in one direction produces a redundant in both.

It also produces a larger answer. The symmetric frame’s largest moment was 3.95 kN·m; this one’s is 6.74, from the same 10 mm on a frame that is less material. Shortening one column has stiffened the frame against the movement — the short column is the stiff one, and the stiff one is the one that resists — and stiffness is what a settlement is paid for in.

That is the settlement case’s version of the rule this whole essay is about. Under load, moving material toward one side of a frame moves moment toward it and is usually what a designer wants. Under a settlement, the same move increases the total, because there is no total to share out: the moments are manufactured by the deformation and the only thing that sets their size is how hard the frame resists.

Where the case comes from #

A settlement case is not an exotic one and it is worth naming the situations that produce it, because they are all situations in which nobody thinks about the frame.

A new building beside an old one. The new foundation loads the ground the old footings sit on, and the old building settles on the side nearest the new one. The differential is across an existing frame that was never analysed for it.

A footing on a made ground boundary. One foot on rock and one on fill is the classic case and the worst, because the differential is not a settlement of one foot but the whole of the fill’s compression.

Dewatering, or a tree. Shrinkable clay loses volume when it dries and gains it when it wets, and a mature tree removed from beside a building lets the clay swell — which is a settlement with the sign reversed, doing exactly the same thing to the frame.

Tunnelling and adjacent excavation. A trough is exactly a differential settlement field, and a propped excavation beside an existing frame imposes one on it as the props are installed and removed.

And thermal movement, which is the same arithmetic. A frame whose beam expands relative to its columns is a frame with an imposed deformation, and the analogy handles it the same way: the released structure’s cut opens by the free expansion, and the section is loaded by that instead of by a moment diagram. An integral bridge is that case every day of its life rather than once, and the movement nobody applied is the general statement of it.

The same arithmetic in other structures #

A continuous beam over a settling pier is the version most people meet first, and its behaviour is the one-dimensional shadow of this: the moment over the settled support is $6EI\delta/L^2$ for a symmetric two-span beam, with the same $EI$ in front and the same absence of any load. The support that moved is that calculation.

An arch whose abutments spread is the case where the sign is reversed and the consequence is larger. An arch resists a spreading of its springings, so the imposed displacement is horizontal rather than vertical, and it enters through $\Delta u_x$ and therefore through $H$ — which reduces the thrust rather than adding a moment. A masonry arch answers a spreading abutment by hinging, which is a controlled failure that leaves the arch standing and is the reason so many old arches have cracks at the haunches.

A tied arch removes the case entirely by taking the thrust in a tie instead of the ground, which is the structural form that exists because of it.

In all three the pattern is the same and it is the one the analogy makes visible: an imposed displacement is resisted in proportion to stiffness, and a structure that resists nothing carries nothing.

Which free body produced the number #

The free body is the frame cut at mid-span of its beam, with the three redundants applied on a rigid arm at the elastic centre. The three conditions are that the cut opens by exactly the relative displacement the settlement imposes on the released structure — which for a foot going straight down is a translation of $\Delta$ and nothing else, because the released structure is two cantilevers and one of them simply goes down with its foot.

The check is the flexibility route on the whole frame with no release at the cut: the settling restraint alone is removed, a unit force is applied there, the resulting deflection is read from the displacement vector, and the answer is scaled by the force that produces exactly 10 mm. The two agree to the precision of the arithmetic at all five stations, and on the reaction as well.

What the picture cannot show #

Time. The settlement is imposed instantaneously here. A real one happens over months or years, and a concrete frame creeps while it does — which relaxes the moments by a factor that depends on how fast the ground moves relative to how fast the concrete does. A settlement applied slowly to a creeping frame can be worth a third of what this calculation says.

Cracking. The linearity that lets the load case and the settlement case be added is the linearity a cracked section does not have. A frame that has cracked at its knees has a lower effective $EI$ there, which sheds settlement moment into the uncracked parts and reduces the total — so the settlement case is partly self-limiting in a way the load case is not.

The foundation. The foot is given a displacement and asked to hold it. A real footing is a spring: the frame’s resistance to the settlement pushes back on the footing, which settles a little less, which reduces the moment. Modelling the foot as rigid and then imposing a displacement on it is conservative and is not the same problem.

Whether the ground is done. Ten millimetres today is a statement about ten millimetres; the relevant number is the final differential settlement, and it is an estimate with a factor of two in it.

The assumption that carries the most #

That the settlement is known. Every number here is proportional to $\Delta$, so the entire calculation is an amplifier for a quantity nobody can measure in advance. A settlement estimate good to a factor of two produces a moment good to a factor of two, and there is no averaging anywhere in the chain to soften it.

That is a genuinely different epistemic position from a load case, and it is why the two are treated differently in practice rather than merely computed differently. A load is known to perhaps twenty per cent and has a factor applied to cover the rest. A differential settlement is known to a factor of two, cannot usefully be factored, and is instead handled by making the structure insensitive to it — by articulating it, by pinning the feet instead of fixing them, or by accepting that it will crack and detailing for the crack.

Pinning the feet is the largest of those moves and the analogy prices it in one step. A pinned foot is a zero-stiffness end, so its band on the analogous column is infinitely wide; $A$ and $I_y$ both grow without limit, and $V = EI\Delta/I_y$ goes to zero. A three-pinned frame develops no settlement moments at all, which is the reason determinate structures are chosen for bad ground and the reason they are not chosen anywhere else.

Still open: the frame with two cells #

Everything the analogy has done, on straight frames, on arches and now on a support that moved, has rested on one number: three. A closed frame cut once has three redundants; a section has three stress resultants — a direct stress and two bending stresses; and the analogy works because those two threes are the same three.

Add a second bay and the frame has six. There is no section with six stress resultants, so the correspondence that the whole method is built on has nothing to correspond to. Whether a two-cell frame can be handled by two analogous columns with something passed between them, whether the analogy degenerates into an ordinary flexibility calculation dressed up, and what exactly the number three was a property of — the frame, the cut, or the plane it is drawn in — is the question after this one.