The axes that have to be turned first

Every frame the column analogy has been applied to so far has been symmetric about a vertical line. The single-bay portal had two columns of the same height and the same stiffness; the fixed arch had a crown on its axis of symmetry. In both, the elastic centre was found, the three redundants came out of three divisions, and the construction looked like a method.

It was a method with a condition on it that never had to be stated, because it was never violated. The condition is that the analogous column’s principal axes are the horizontal and the vertical — and a symmetric section’s are, so nobody had to say so.

Make one column shorter than the other and it stops being true.

What the section acquires #

The three quantities the analogy needs from the frame are its area at a width of $1/EI$, its centroid, and its second moments about that centroid. For a symmetric frame the two second moments are all there is. For an unsymmetric one there is a third:

$$I_{xy} = \int (x - \bar x)(y - \bar y)\,\frac{ds}{EI},$$

and it is zero only when the section has an axis of symmetry. On a portal with two columns of the same height the two legs contribute equal and opposite amounts and the beam contributes nothing, because the beam lies at one height and the first moment of a symmetric strip about its own centroid is zero. Shorten one leg and none of those cancellations survive: the centroid moves sideways, the beam’s first moment about it is no longer zero, and the two legs are no longer mirror images.

The consequence is not that the elastic centre moves. It is that finding it is no longer enough.

Where the coupling comes from #

The three compatibility conditions have not changed. Cut the frame at the crown, apply the three redundants on a rigid arm at the elastic centre, and require that the cut does not open: no relative horizontal movement, no relative vertical movement, no relative rotation. With $M = M_s + M_0 + H(y - \bar y) + V(x - \bar x)$ those three conditions are

$$\int M \,dA = 0, \qquad \int M (y - \bar y)\,dA = 0, \qquad \int M (x - \bar x)\,dA = 0,$$

where $dA$ stands for $ds/EI$. The first still gives $M_0 = -P/A$ on its own, because the first moments about a centroid are zero whatever the shape. The other two give

$$H I_x + V I_{xy} = -Q_x, \qquad H I_{xy} + V I_y = -Q_y,$$

with $Q_x = \int M_s(y-\bar y)dA$ and $Q_y = \int M_s(x-\bar x)dA$. They are coupled through $I_{xy}$ and they cannot be solved one at a time. Set $I_{xy}$ to zero and they fall apart into $H = -Q_x/I_x$ and $V = -Q_y/I_y$, which is the symmetric construction — and which is the shape the method has in every worked example anybody learns it from.

The same coupling is the whole of unsymmetric bending. A steel angle loaded straight down deflects sideways because its principal axes are not the ones its load is applied about, and a section with a product of inertia has a neutral axis that is not perpendicular to the plane of loading. The analogous column is a section like any other and it does exactly the same thing. What is different, and what makes it hard to notice, is that the “stress” being computed is a bending moment in a frame and the “load” is a released moment diagram, so nothing about the arithmetic looks like a section at all.

What the shortcut costs #

Thirteen per cent is a large error for a method whose point is exactness, and where it falls is worse than how large it is.

The left foot is understated by 39 per cent — 17.0 against 27.8 — and the left foot is where a fixed-base portal is designed. The knees are treated more gently, and the crown is almost unaffected. That distribution is not luck: the error enters through $V$, the redundant shear, whose contribution to the moment is $V(x - \bar x)$ and is therefore largest at the ends of the frame and zero on the line through the elastic centre.

So the shortcut is at its worst exactly where the answer is used, and at its best in the middle of the beam where a moment of 38 against 38 looks like agreement. An engineer checking one number against a handbook would check the crown.

The three-way comparison is what makes the claim a measurement rather than a statement of confidence. Two of the three answers agree to a tenth of a per cent and the third does not, and the two that agree were obtained by completely different routes: one integrates a released moment diagram over a fictitious section, the other assembles a stiffness matrix for twenty-six members and inverts it. A shortcut that disagreed with only one of them would be an open question.

How lopsided is lopsided enough #

The curve has no threshold in it, which is the useful thing about it. There is no ratio below which the coupling can be neglected and above which it cannot; the error rises smoothly from zero and is roughly twice the tilt angle in degrees.

That gives a rule worth carrying: six degrees of tilt is ten per cent of the answer. Six degrees is not a lopsided frame. It is a column-height ratio of about 0.67 — a five metre column beside a three-and-a-third metre one, which is an ordinary building on a slope, an ordinary portal with a mezzanine at one end, an ordinary shed whose eaves step. At a ratio of 0.72, which is a frame nobody would look at twice, the error is still 7.8 per cent. Nothing on the drawing looks unusual and nothing in the calculation announces itself.

Height is not the only way in. Two columns of the same height with different second moments do it too, and so does a beam whose stiffness differs from either column’s in an unsymmetric arrangement.

That case is the one most likely to be missed, because the drawing is symmetric. A portal with one column stiffened — for a crane rail, for a wind post, because one side carries a mezzanine — is a frame nobody would call unsymmetric, and its analogous column is tilted further than the visibly lopsided one above it.

Where the axes actually are #

The tilt is the ordinary construction of principal axes, applied to a section made of $1/EI$:

$$\tan 2\alpha = \frac{2I_{xy}}{I_x - I_y},$$

and the two principal second moments are $\tfrac12(I_x + I_y) \pm \sqrt{\tfrac14(I_x-I_y)^2 + I_{xy}^2}$ — Mohr’s circle for second moments, on an object that is not a cross-section of anything.

It is worth noticing which way the tilt goes and why it is small even when $I_{xy}$ is large. For this frame $I_x = 31.6$ and $I_y = 168.0$: the section is five times stiffer about the vertical axis than about the horizontal one, because the analogous column is 8 m wide and about 4 m tall. A large difference between $I_x$ and $I_y$ divides a large $I_{xy}$ down, so the tilt is 7.5 degrees rather than 30. A tall narrow frame — a portal whose columns are much longer than its span — has $I_x$ and $I_y$ closer together and tilts much further for the same asymmetry, which is the case the rule of thumb above is least safe for.

There is a second route to the same answer and it is the one worth using if the arithmetic is being done by hand. Rotate the axes to the principal ones, compute $Q_1$ and $Q_2$ about them, divide by $I_1$ and $I_2$, and rotate the two resulting forces back. It is two divisions again, with a rotation on each end — and it is exactly what a designer does with an angle in biaxial bending, for exactly the same reason.

The whole of it, once, by hand #

For the frame with $L = 8$ m, $h_L = 5$ m, $h_R = 3$ m, $w = 10$ kN/m and all three members at $EI = 1$:

The analogous column is the centreline at unit width, so its area is the total length, $A = 5 + 3 + \sqrt{8^2 + 2^2} = 16.25$. Its centroid comes out at $\bar x = 3.508$ m, $\bar y = 3.077$ m — pulled left and down by the taller left column, which carries more of the section’s area than the short right one does. About that centroid,

$$I_x = 31.57, \qquad I_y = 168.04, \qquad I_{xy} = -18.38 .$$

The release is a cut at mid-span, so $M_s = -wL^2/8 = -80$ kN·m down both columns and $M_s(x) = -5(4-x)^2$ along the beam. The three integrals of it over the section are

$$P = -859.9, \qquad Q_x = +406.1, \qquad Q_y = +216.6 .$$

The first redundant is a division: $M_0 = 859.9/16.25 = 52.93$ kN·m. The other two are a two-by-two,

$$\begin{bmatrix} 31.57 & -18.38 \\ -18.38 & 168.04 \end{bmatrix}\begin{bmatrix} H \\ V \end{bmatrix} = \begin{bmatrix} -406.1 \\ -216.6 \end{bmatrix} \;\Rightarrow\; H = -14.54, \quad V = -2.879 .$$

Dropping $I_{xy}$ turns that into $H = -406.1/31.57 = -12.86$ and $V = -216.6/168.0 = -1.289$: the thrust is understated by 12 per cent and the redundant shear by 55.

Then the left foot, at $(0, 0)$:

$$M = -80 + 52.93 + (-14.54)(0 - 3.077) + (-2.879)(0 - 3.508) = -80 + 52.93 + 44.74 + 10.10 = 27.77\ \text{kN·m},$$

against the shortcut’s $-80 + 52.93 + 39.58 + 4.52 = 17.03$. Both of the coupled terms are wrong and both are wrong the same way, so the two errors add rather than partly cancelling — which is why the foot is out by 39 per cent when neither redundant is out by that much on its own.

Where this turns up, and where it is worst #

The frame with two column heights is the textbook case and it is not the common one. Four arrangements produce the same coupling and only one of them looks lopsided:

A building on a slope. Columns founded at two levels, beam horizontal, everything else identical. This is the case in the figures and it is the one an engineer would notice.

A portal with one column stiffened. Same heights, same span, one column a heavier section for a crane rail or a wind post. The drawing is symmetric and the analogous column is tilted further than the visibly lopsided one, because the section’s width is $1/EI$ and stiffening a member makes its band narrower.

A pitched portal with an off-centre ridge. The beam is two segments and the vertex is not at mid-span, so the section has a kink in a place that breaks its symmetry.

A frame with one fixed foot and one pinned. A pin is a zero-stiffness end, which in the analogy is a band of infinite width, and the section becomes unsymmetric in the most extreme way available. That case is usually handled by a different release rather than by the analogy, and the reason is the one this paragraph names.

Where it is worst is the tall narrow frame, and the reason is in the arithmetic of the tilt rather than in the asymmetry. The tilt is set by $2I_{xy}/(I_x - I_y)$, and for the frame drawn here $I_y$ is five times $I_x$ — an 8 m wide section only 4 m tall — so a large product of inertia is divided down and the axes tilt 7.5 degrees. A portal 4 m wide and 8 m tall has $I_x$ and $I_y$ much closer together, and the same asymmetry tilts the axes much further. Multi-storey frames, bridge piers and lift shafts all live in that region.

Which free body produced the number #

The free body is the frame cut at mid-span of its beam, leaving two bent cantilevers each fixed at its own foot. The released moment is $-wL^2/8$ constant down each column and $-w(L/2-x)^2/2$ along the beam, with $x$ measured horizontally from the left knee — the same release the symmetric construction used, because the release does not have to respect a symmetry the frame does not have.

The three redundants are a couple, a horizontal force and a vertical force, applied on a rigid arm at the elastic centre. The three conditions are that the cut does not open. The check is a stiffness solution of the same frame with no release in it at all, and it agrees to 0.10 per cent of the free moment.

The generalisation #

The pattern here is older and wider than the column analogy, and stating it in general is what makes it findable elsewhere.

Any method that solves a redundant structure by choosing a good set of releases is diagonalising a matrix. The force method writes $\mathbf{f}\,\mathbf{X} = -\mathbf{\Delta}$, with $\mathbf f$ the flexibility matrix of the chosen releases; choosing what to take away is choosing that matrix, and a good choice is one whose off-diagonal terms are small. The elastic centre is the choice that makes them exactly zero — for a symmetric structure. For an unsymmetric one it zeroes two of the three and leaves the third, and $I_{xy}$ is that third.

So the analogy’s elegance was never that it avoided a matrix. It was that a particular choice of where to apply the redundants made the matrix diagonal, and that choice is available to any method: moment distribution converges fast for the same reason, because passing a moment round a joint is an iteration on a matrix that is nearly diagonal already.

And the second half of the pattern is the one to carry. A method demonstrated only on symmetric examples is a method with a hidden condition, and the condition is invisible precisely because the examples were chosen to be easy. There is no step in the symmetric construction at which $I_{xy}$ is set to zero — it simply is zero, so nobody writes it down, and the reader learns a three-division method rather than a two-by-two with a term that vanished. Loading an angle straight down has the same shape: every beam in a textbook is doubly symmetric until one is not, and the first unsymmetric one is the one that deflects sideways.

The test that would have caught it is the one worth building into any hand method: run it on a case the symmetry does not cover, and check it against something that shares no arithmetic with it. Here that is a stiffness solution, and it takes 0.1 per cent of the free moment to say which of two plausible answers is the right one.

What the picture cannot show #

Sway. The load here is vertical and the frame is fixed at both feet, so the horizontal redundant is a thrust rather than a sway force. A lateral load on the same frame is handled by the same three conditions and the same $2\times2$, and the distribution of it between two columns of different heights is a stiffness question with a cube in it.

Axial shortening. Every integral is a bending flexibility. The columns shorten under their own share of the load and the beam shortens under the thrust, and none of that is in the section.

The knee. The analogous column is a centreline, so the frame’s corners are points. A real knee is a region a metre across with a haunch in it, and the analogy’s own answer for what a haunch does is the one the straight member gave.

Whether the beam is straight. The beam here slopes because the columns differ; a pitched portal with a ridge is a different section again, with two beam segments and a vertex, and nothing in the construction objects.

The assumption that is easiest to carry across #

That the released structure’s moment diagram is the one written above. It is, for the release chosen — but the choice of release is free, and a different one gives a different $M_s$, a different $P$, $Q_x$ and $Q_y$, and the same final answer. That invariance is worth testing rather than assuming when the arithmetic is done by hand, because it is the one property of the method that no single calculation can demonstrate.

The other assumption is the one this essay is about, and it is worth stating in the form that makes it visible: the symmetric construction is not a simplification of the general one, it is the general one with a term that happens to vanish. A method learned only on symmetric examples has a missing term in it that never announces itself, because the answer it gives is smooth, plausible, and wrong by thirteen per cent at the station that governs.

Still open: the support that was never loaded #

Every load on this page and on the two before it has been a force. The analogy’s three “stresses” came from loading the analogous column with a released moment diagram, and that diagram came from the frame’s loads.

A frame can be bent by something that is not a load at all. One foot of a portal settles by ten millimetres; the frame is three times redundant, so the settlement is resisted, and moments appear throughout with no load anywhere on the structure. In the analogy that case has a form so simple it is suspicious: an imposed displacement is a concentrated load on the analogous column, applied at the point that moved, and the answer scales with $EI$ rather than with anything applied. What that concentrated load is, where exactly it goes, and why a stiffer frame is punished rather than rewarded for its stiffness, is the question after this one.