The brace on the wrong flange

A column’s brace has to answer one question: how stiff. Reach the ideal stiffness and the column behaves as two shorter columns; go past it and nothing further is bought, because the column has stopped using the brace.

A beam’s brace has to answer two, and the second one is not about stiffness at all. Lateral-torsional buckling moves the two flanges in opposite directions, so where the restraint sits in the depth of the section decides whether it is fighting the buckle or standing next to it.

On the beam below, a restraint of 447 kN per metre on the compression flange does the whole job. The identical restraint on the tension flange buys 7%, and increasing it by a factor of a hundred buys 14%.

Which free body produced the number #

There is no free body here; there is an energy balance, because buckling is a question about whether a shape costs energy or releases it.

Take a simply supported beam under uniform moment and give it a lateral displacement $u$ and a twist $\varphi$, each as a sum of half-waves:

$$u = U_1 \sin\frac{\pi x}{L} + U_2 \sin\frac{2\pi x}{L}, \qquad \varphi = \Phi_1 \sin\frac{\pi x}{L} + \Phi_2 \sin\frac{2\pi x}{L}$$

The strain energy has three terms — minor-axis bending $EI_z u''^2$, warping $EI_w \varphi''^2$ and St Venant torsion $GJ\varphi'^2$ — and the applied moment does work $-M\!\int u''\varphi\,dx$. A spring of stiffness $k$ at midspan, sitting a height $a$ above the shear centre, adds $\tfrac12 k(U_1 + a\Phi_1)^2$.

Under uniform moment the two half-waves do not couple to each other, so the whole problem is two 2×2 determinants. With no spring the first gives

$$M_{cr} = \frac{\pi}{L}\sqrt{EI_z\left(GJ + \frac{\pi^2 EI_w}{L^2}\right)}$$

which is the standard result, exactly, and 143.4 kNm for the section drawn.

The spring appears only in the first half-wave, because $\sin(2\pi x/L)$ is zero at midspan. So as the brace stiffens, the first mode is driven up and the second is not, and the answer climbs to a plateau at the second mode’s value and stops.

The plateau is the half-length beam, and it is not twice #

The second half-wave’s critical moment is the same expression with $2\pi/L$ in place of $\pi/L$, which is exactly the unbraced formula for a beam of length $L/2$. That is the right answer and it is worth checking against intuition, because intuition gets the size wrong.

Halving the length multiplies the leading $\pi/L$ by two, and multiplies the warping term inside the root by four while leaving $GJ$ alone. So the gain is

$$\frac{M_{cr}(L/2)}{M_{cr}(L)} = 2\sqrt{\frac{GJ + 4\pi^2 EI_w/L^2}{GJ + \pi^2 EI_w/L^2}}$$

which is between 2 and 4 depending on where the section sits between pure torsion and pure warping. For the beam here — $GJ = 2.51\times10^{10}$ against $\pi^2EI_w/L^2 = 2.30\times10^{10}$, almost exactly balanced — it is 3.12.

A section with no warping stiffness at all, a hollow one for instance, would give exactly 2. A deep thin-flanged plate girder, whose behaviour is nearly all warping, approaches 4. The gain from a brace is a property of the section.

Why the tension flange is useless #

The buckled shape has $u$ and $\varphi$ of opposite sign — that is what makes the moment’s work term negative and the buckle worth doing. So the section is rotating about a point somewhere near the tension flange, and a brace there is very nearly on the axis of rotation.

A restraint on the axis a body is rotating about restrains nothing. It has almost no displacement to resist, so it develops almost no force, so it stores almost no energy, so it does not change the load at which the energy balance tips.

Sweeping the brace height on the beam here, with the compression flange’s ideal stiffness applied throughout:

Nearly a factor of three across the depth of one section, from a restraint that never changed.

The span decides how much a brace is worth #

The plateau and the unbraced value both fall with span, and they do not fall together — so the value of a brace changes with the beam it is on.

The gain falls as the beam lengthens, which is the opposite of what most people expect. The reason is the warping term: it goes as $1/L^2$ inside the root, so on a short beam it dominates and halving the length is worth close to four, while on a long beam $GJ$ dominates and halving is worth close to two.

The ideal stiffness moves the other way and much faster — 1,062 kN/m at 6 m against 229 at 10 m, a factor of 4.6 across a factor of 1.7 in span. A short beam needs a much stiffer brace and gets more for it.

What a real brace is, and why the answer changes #

The model above puts a lateral spring at a height. Real bracing comes in two forms and they map onto that model differently.

A lateral brace attached to the compression flange — a purlin, a tie, a plan brace — is exactly the model: a spring at $a = +h/2$.

A torsional brace — a cross-member framing into the web, a stiffener connected to a slab, a moment connection to a secondary beam — restrains $\varphi$ directly rather than $u$. It is a rotational spring, not a lateral one, and it enters a different term of the same determinant. The interesting thing is that it does not care about height at all: a torsional restraint at the shear centre is exactly as good as one anywhere else, because it is resisting the rotation rather than a displacement caused by it.

Which gives the practical division. Where the compression flange is accessible — the top flange of a simply supported beam under gravity load — brace it laterally, and it is cheap. Where it is not — the bottom flange of the same beam over a support, in a continuous system, where the slab is on the wrong side — the answer is a torsional restraint, and that is why full-depth stiffeners and knee braces appear at the supports of continuous beams and nowhere else along them.

The stiffness needed, and how little it is #

The ideal stiffness for the beam here is 447 kN/m at the compression flange. That is a small number: a 3 m long tie of 500 mm² area has an axial stiffness of $EA/L = 35{,}000$ kN/m, seventy-eight times more than required.

That is the usual finding and it is the same finding as for a column: bracing is a stiffness requirement and the stiffness required is trivially available, so the design question is almost never whether the brace is stiff enough. It is whether the brace is connected stiffly enough, and whether the thing it is attached to at the far end is going anywhere.

Half the ideal stiffness is worth a good deal less than half the benefit: 295 kNm rather than 447, so 50% of the stiffness buys 50% of the gain. Quarter stiffness gives 220, which is 25% of the gain. The relationship is very nearly linear right up to the plateau and then stops dead — which is the same shape a column’s brace curve has and comes from the same crossing of two modes.

The same argument at the support of a continuous beam #

Everything so far has been about a simply supported beam whose compression flange is on top. Turn the moment over and the whole geometry inverts.

Over the interior support of a continuous beam the moment is hogging, so the bottom flange is in compression. It is the flange furthest from the slab, closest to nothing, and generally the one nobody has arranged to hold. Meanwhile the slab is bolted to the top flange, which is in tension — the flange this page has just spent several hundred words showing is worth almost nothing to brace.

So a continuous beam with a composite slab has excellent bracing along the parts of it that need none and none at all along the part that does. The remedies are the torsional ones: a full-depth stiffener that makes the slab’s restraint of the top flange into a restraint of the whole section, or a knee brace from the bottom flange up to the slab.

That geometry is why continuous composite beams have detailing at their supports that simply supported ones do not, and why the length between the support and the point of contraflexure is the unbraced length that governs.

Where the model stops #

Uniform moment is the worst case and it is not the usual one. A beam under a uniform load has its largest moment at midspan and less elsewhere, so less of its length is at the moment that is trying to buckle it — the critical moment is higher by a factor between 1.1 and 1.7 depending on the shape of the diagram. Every number here is conservative for that reason.

The load’s height is not in the model. A load applied at the top flange is destabilising: as the section twists, the load moves sideways with the flange and its line of action develops a lever arm about the shear centre. A load at the bottom flange is the opposite. That is a second height question, independent of the brace’s, and on a beam carrying a slab on its top flange it can be worth 20%.

And the brace is elastic and perfect. A real brace also has to carry a force — small, of the order of 1–2% of the flange force, and coming entirely from the beam’s initial out-of-straightness. The eigenvalue model says the force is zero, because a perfect beam does not push on its brace until it buckles, and that is exactly why the force has to be estimated from the imperfection instead.

What the pictures cannot show #

The two-term Ritz solution has exactly two modes in it, so the plateau is exactly the second one. A real beam has infinitely many, and near the plateau the true answer is a mode that is neither of them — the brace point is not quite a node, and the beam is not quite two independent halves.

Nor can the drawings show the twist. The buckled beam moves sideways by some tens of millimetres and rotates by a few degrees, and all of the figures on this page are plots rather than pictures of a beam.

The assumption the figure rests on #

The brace is a spring in one direction, at one point, with no mass and no strength limit. Every real one is a member with two ends, and the far end is attached to something that is also moving. A row of beams braced to each other and to nothing else is a row of beams that will all buckle together in the same direction — the braces are perfectly stiff relative to each other and offer the system no restraint at all, which is the failure mode that plan bracing exists to prevent and that this model, with its immovable spring, cannot represent.

Bracing during erection is a different problem with the same equations #

Every number above is for a finished beam. The beam that most needs bracing is the one that has just been lifted into place, and its situation differs in three ways that all point the same direction.

Its unbraced length is the whole span, because nothing has been attached yet. Its load is only its own weight, which is small — so the demand is small too, and that saves it. And its bracing, when it arrives, is temporary: a tie to the beam next door, which is also unbraced, and which will happily buckle in sympathy.

The third of those is the one that bites. Two beams tied to each other are stiff against each other and free to move together, and the mode that costs least is the one where they do. That is not a stiffness failure — the tie can be as stiff as anybody likes — it is a failure to anchor the bracing system to anything that is not moving.

The ladder from here #

Later rungs on this anchor: torsional bracing developed properly, with the rotational spring in the determinant and the finding that its ideal stiffness is proportional to the square of the moment rather than the first power. Load height, and the two-parameter family of critical moments that follows. Multiple braces, where the plateau is the $n$th mode and the ideal stiffness rises with the number of braces rather than falling. The brace force from an assumed imperfection, and where 2% comes from. Continuous restraint from a deck, where the spring becomes a foundation modulus and the mode has a wavelength of its own. And the case where the beam braces the brace: a row of beams whose plan bracing has to be anchored somewhere.