The corner that moves most

A storey of a building is pushed sideways by wind or by its own inertia. The push is shared out among the walls, frames and cores that resist it, and the obvious rule — each takes a share in proportion to its stiffness — is correct exactly once: when the push passes through a particular point in the plan.

That point is the centre of rigidity, the stiffness-weighted centroid of everything resisting. It is not the centre of the plan, it is not the centre of mass, and the distance between it and the line of the push is an eccentricity that applies a torque to the whole storey.

Where the centre of rigidity is #

The definition follows from insisting the storey does not rotate. If every resisting element deflects by the same amount $u$, each carries $k_i u$, and the resultant of those forces acts at

$$x_R = \frac{\sum k_i x_i}{\sum k_i}$$

taken over the elements resisting in the direction of the push. That is a centroid with stiffness in place of area, which is exactly the arithmetic of a centroid of area with a different weighting — and the reason it is worth naming separately is that stiffness is a much more concentrated quantity than area. A shear wall’s stiffness goes as the cube of its length, so a core four times the length of a façade wall is not four times as stiff but very much more, and the centre of rigidity is dragged toward it.

For the plan drawn: a core of stiffness 400 at $x = 6$ m and a façade wall of 100 at $x = 30$ m give

$$x_R = \frac{400 \times 6 + 100 \times 30}{500} = 10.8\;\text{m}$$

while the mass, spread over the floor, sits at 15.0 m. The eccentricity is 4.2 m — a seventh of the plan width, from an arrangement that would look entirely ordinary on a drawing.

Which free body produced the number #

Take the whole storey as one rigid body, with the walls as springs beneath it.

Move it by a translation $u$ and a rotation $\theta$ about the centre of rigidity. Each element’s displacement is $u + \theta d_i$, where $d_i$ is its distance from that centre measured perpendicular to its own direction of action, so each carries $k_i(u + \theta d_i)$.

Two equilibrium equations follow. Summing forces gives $V = u\sum k_i$, and summing moments about the centre of rigidity gives $Ve = \theta \sum k_i d_i^2$, since the direct terms have no moment about that point by its own definition. Write $J_R = \sum k_i d_i^2$ — the torsional stiffness of the plan — and each element carries

$$V_i = \underbrace{\frac{V k_i}{\sum k_i}}_{\text{direct}} \;+\; \underbrace{\frac{V e\, k_i d_i}{J_R}}_{\text{torsional}}$$

Note what is in $J_R$: every element, including those at right angles to the push. Walls running east–west contribute nothing to resisting a north–south force and a great deal to resisting the twist it causes, which is the first thing this arrangement gets wrong in an intuition trained on beams.

The identical arithmetic appears three orders of magnitude smaller in a bolt group under an eccentric load, and in a weld group as its continuous version. Every one of them is the same two terms: a direct share, and a term proportional to distance from a centroid.

What makes the plan case worse than either is that the distances are storey-sized rather than millimetre-sized, and that nobody draws the centroid.

Moving the force is where the torque comes from #

The step that produces the torque is one this collection has already made in another setting, and naming it makes the whole arrangement obvious.

A storey force acts through the mass centre. To use the proportional rule it has to act through the centre of rigidity. Moving a force to a new point is free only if a couple is added, and the couple is the force times the distance moved. So the eccentric push is exactly equivalent to a concentric push plus a torque of $Ve$ — not approximately, not as a modelling assumption, but as an identity.

Which means the two terms in the wall-force expression are not two mechanisms. They are one force, decomposed at a point chosen so that the decomposition is easy, and the point that makes it easy is the one about which the direct shares have no moment.

The far wall pays #

Read the numbers off the plan above, with the mandatory accidental eccentricity of 5% of the plan width added to the natural 4.2 m:

The core carries the most and is relieved by the torsion, because it sits on the same side of the centre of rigidity as the push. The wall in trouble is the façade wall: designed for 200 kN by the proportional rule, asked for 396 — 98% more than its direct share.

That inversion is the practical content of the subject. The element that fails a plan-torsion check is almost never the one carrying the largest force; it is the small one at the far end that nobody was watching, and the reason it is in trouble is that its direct share was small enough for the torsional term to double it.

And the two walls at right angles, carrying 55 kN each with nothing applied along them at all. They appear in no load path anybody drew.

The symmetric plan that twists worst #

The quantity that separates the two plans is the torsional radius

$$r = \sqrt{J_R / \textstyle\sum k_i}$$

which is a length: the distance at which all the stiffness would have to sit to give the plan the torsional stiffness it has. It is the plan’s radius of gyration, computed with stiffness in place of area, and it plays exactly the part $\sqrt{I/A}$ plays for a column.

Compare it with the plan’s own radius of gyration of area, $l_s = \sqrt{(L_x^2 + L_y^2)/12}$, and the criterion falls out. If $r > l_s$ the plan is torsionally stiff and a small eccentricity produces a small twist. If $r < l_s$ it is torsionally flexible, and the twist governs everything.

The second row is the reason accidental eccentricity is mandatory rather than advisory. A plan with a central core has no calculated eccentricity, so a designer computing the torsional case from the natural eccentricity alone would find no torsion at all — and the building would still twist, because the mass is never quite where it was assumed and the stiffnesses are never quite what was calculated. Requiring 5% of the plan dimension as an eccentricity turns an unanswerable question into a load case, and on a torsionally flexible plan that load case dominates.

Stiffening it makes it worse #

The arithmetic is short. Adding $\Delta k$ at $d = 0$ changes $\sum k$ and leaves $J_R$ alone. The translation falls as $V/(\sum k + \Delta k)$, the rotation $Ve/J_R$ does not move, and the ratio of edge displacement to centre displacement is

$$1 + \frac{\theta d_{edge}}{u} = 1 + \frac{V e\, d_{edge}}{J_R} \cdot \frac{\sum k + \Delta k}{V}$$

which rises linearly with the stiffness added. The absolute displacement at the edge does fall — the building is genuinely stiffer — but the rotation does not, and the rotation is what cracks the cladding, jams the lift guides and drives the second-order effects at the far corner.

The variable is position, not quantity. A wall of stiffness 100 at the far end of the plan contributes $100 \times 19.2^2 = 36{,}900$ to $J_R$, while a wall of stiffness 400 at the centre contributes nothing at all. Four times the material, none of the effect.

In elevation the same question is what a tall building’s lateral system is for, and in plan it is this.

A torsional displacement is not merely an aesthetic problem either. Second-order effects grow with the displacement they act on, so the far corner of a twisting plan carries an amplification the centre never sees, and it carries it on the columns least able to be checked for it.

The table above is four load cases #

The two rows worth re-reading are the first two, because they were not computed with the same eccentricity and the difference is the whole of how the check is run.

The accidental eccentricity is a displacement of the mass centre, and nothing says which way it has moved. It is $\pm 1.5$ m on a 30 m plan, applied in whichever sense is worse for the element being checked. For the façade wall, which the torsion loads, that means adding it: $e = 4.2 + 1.5 = 5.7$ m. For the core, which the torsion relieves, the worst case is the one that relieves it least: $e = 4.2 - 1.5 = 2.7$ m.

Run both through $V_i = Ve\,k_i d_i / J_R$, with $J_R = 55{,}800$ for this plan:

$$\text{façade: } \frac{1000 \times 5.7 \times 100 \times 19.2}{55{,}800} = +196\ \text{kN}$$

$$\text{core: } \frac{1000 \times 2.7 \times 400 \times 4.8}{55{,}800} = -93\ \text{kN}$$

Note that $k_i d_i$ is 1,920 for both — it must be, since $\sum k_i d_i = 0$ is the definition of the point they are measured from, and there are only two elements to share it. The two torsional shares differ by a factor of two solely because the two eccentricities do.

A calculation that applies one signed $e$ to every element at once gets the core wrong. At $e = 5.7$ the core reads $800 - 196 = 604$ kN instead of 707 — seventeen per cent light, in a spreadsheet whose façade wall is correct and whose total is correct and which therefore looks finished. The error is invisible in every sum on the page, because the load that went missing from the core went into the façade wall, where it also belongs.

So the four-row table is four separate analyses sharing a plan, and on a plan with walls in both directions and two senses of push it is sixteen. That count is why the subject is done by machine, and why the check on the machine is the arithmetic above run once by hand.

The strong direction is the vulnerable one #

Push the same plan east–west instead, and almost every number changes — including the one the plan is judged by.

The centre of rigidity moves: for the east–west push it is set by the two façade walls, $y_R = (60 \times 0 + 60 \times 18)/120 = 9.0$ m, which is the plan’s own centre. The natural eccentricity in that direction is zero, so only the accidental 5% of 18 m applies, $e = 0.9$ m.

But $J_R$ does not change. It is a property of the plan and every wall is in it, whichever way the push goes — so the torsional radius is computed with the same numerator and a different denominator:

$$r_y = \sqrt{\frac{55{,}800}{500}} = 10.56\ \text{m}, \qquad r_x = \sqrt{\frac{55{,}800}{120}} = 21.6\ \text{m}$$

Against $l_s = 10.10$ m, that is $r/l_s$ of 1.05 one way and 2.13 the other. The edge amplification follows: $1 + e\,d_{edge}/r^2$ gives 1.98 north–south and 1.02 east–west. One plan, one set of walls, and a torsional problem in one direction and none at all in the other.

The direction that has the problem is the direction with more resisting stiffness. That reads backwards until it is set beside the previous section’s result and seen to be the same statement: $r^2 = J_R/\sum k$ has the shared torsional stiffness on top and the directional stiffness underneath, so stiffening a plan in one direction — without moving anything outward — divides its torsional radius down. Adding walls near the centre of rigidity does it fastest, since they add to $\sum k$ and nothing to $J_R$.

Which turns the design rule inside out. The question is never how much stiffness a plan has; it is how far out the stiffness sits, and the two are traded against each other by every wall that goes in.

There is a ceiling on the trade, and it is a property of the footprint rather than of the walls. Split a direction’s stiffness into two equal halves and put them at the two edges, $x = 0$ and $x = L_x$: the centre of rigidity is at $L_x/2$, each half sits $L_x/2$ from it, and

$$r = \sqrt{\frac{2 \times (k/2)(L_x/2)^2}{k}} = \frac{L_x}{2}$$

independent of how much stiffness there is. For this plan that is 15.0 m, against $l_s = 10.10$ — so the best torsional radius the footprint allows is $1.49\,l_s$, and the arrangement drawn achieves 10.56, or 70 per cent of it. That is the number worth carrying out of the whole page: a plan’s torsional capacity is fixed by its own dimensions, the design decides what fraction of it is taken up, and no quantity of material buys any of the remainder.

The floor has to deliver the torque #

All of this assumes the storey behaves as one rigid body in plan. Something has to make it one, and that something is the floor.

If the floor is stiff in its own plane relative to the walls — a concrete slab almost always is — the rigid-body assumption holds and the distribution above applies. If it is not, the storey does not have a single rotation, the centre of rigidity stops being a meaningful point, and the walls share the load by something closer to tributary area.

Where the torque ends up #

The torque is resisted within the storey by the walls, and each wall then hands its own share of it downward as a force. That is worth following, because it is where a plan-torsion problem stops being about a floor plate and becomes about a foundation.

A wall carrying 396 kN rather than 200 delivers 396 kN into whatever is under it — a pad, a pile cap, a piece of raft. Over the height of the building those forces accumulate as an overturning moment on that wall’s own foundation, and the wall that was doubled by torsion has its foundation doubled with it. The core, relieved by the same torsion, has a foundation that could have been smaller.

Two consequences follow, and both are the kind that appear late.

The foundation loads are not symmetric even when the plan is. For the central-core plan above, the accidental eccentricity has to be applied in both directions and both senses, so every wall is designed for the worst of four cases and the core’s own foundation carries a torque about a vertical axis that no gravity calculation contains.

And the torque has to cross every interface on the way down. Each floor delivers its share to the walls, each wall delivers to the next storey down, and the base of each wall delivers to the ground. A discontinuity anywhere in that chain — a wall that stops, a core that changes shape, a transfer level — puts the whole storey’s torque through a member that was designed for a vertical load.

Where the model stops #

Every element here is a spring with one stiffness. A shear wall has bending and shear flexibility, a moment frame has neither in the same proportion, and a core has torsional stiffness of its own that this model ignores entirely. A closed core is a hollow box and its own torsion constant can be a large part of $J_R$ — omitting it is conservative and sometimes very conservative.

The centre of rigidity is a storey-by-storey idea and buildings are not. For a multi-storey building with elements of different shapes, each storey has its own centre of rigidity, and they do not lie on a vertical line. The single point drawn here exists exactly when every element’s stiffness varies up the height in the same proportion.

Everything is elastic and first-order. Once walls crack, their stiffnesses change, and they do not change equally — the most heavily loaded cracks first, its stiffness falls, and the centre of rigidity moves during the event. A plan that was torsionally stiff at working load may not be at collapse.

And the accidental eccentricity is a convention. Five per cent of the plan dimension is a number chosen to cover mass that is not where it was assumed, stiffness that is not what was calculated, and a rotational component of ground motion that nobody measures. It is not a calculation and it should not be reported as one.

What the pictures cannot show #

The plan figures draw walls as heavy lines of arbitrary length, because a wall’s stiffness rather than its length is the input. Two walls drawn the same are not the same wall, and the number beside each is the only honest statement in the drawing.

Nor can they show the rotation. A storey twisting by the amounts computed here turns through a few thousandths of a radian, which over 15 m is a few tens of millimetres — invisible at the scale of a plan, and the reason the whole subject is discussed in ratios.

And the amplification figure draws a straight line rising to 2.96, which is a ratio of two displacements and not a ratio of two forces. The forces move differently, because the direct share also changes as stiffness is added.

The ladder from here #

Later rungs on this anchor: the core’s own torsional stiffness, which is a closed thin-walled section and belongs to the same argument as a box girder. The three-dimensional version, where the centres of rigidity of successive storeys do not line up and the building’s torsional behaviour is a global mode rather than a storey property. Torsional irregularity as codes define it — the ratio of maximum to average storey drift — and why that particular measure was chosen. Accidental eccentricity’s origin in rotational ground motion. Torsionally coupled dynamic response, where the sway and twist modes exchange energy. And the design question this all serves: given a plan and a set of walls, where should they go — which is an optimisation with $J_R$ as its objective and architecture as its constraint.