The lacing decides the force it has to carry

A built-up column has a second way to bend, and the lattice that supplies its shear stiffness has to be sized. The question this rung is about is what force to size it for.

The hero is the column: 24 m, two chords 4,000 mm² each at 800 mm centres, single lacing of 300 mm² diagonals at 45° in two planes. Its Euler load is 4,629 kN and its critical load is 3,832 kN — 83 per cent, and an effective slenderness of 66 against a geometric 60.

Now push 2,000 kN through it and ask what the diagonals carry.

A straight column has no shear at all #

Take the column as perfectly straight, load it axially through both chords, and cut it at mid-height. What crosses the cut is 2,000 kN of axial force and nothing else. No shear, no moment, and therefore no force in any diagonal.

That is not a curiosity; it is the whole difficulty. The lacing exists to supply a shear stiffness that has no shear to resist, and the shear it does resist arrives entirely from the column not being straight.

So the calculation begins with an assumed crookedness. Take an initial bow of $L/500$ — 48 mm on this column — which is the fabrication tolerance a built-up member is made to, and is a number from a workshop rather than from mechanics.

The bow puts the axial load at an eccentricity, the eccentricity makes a moment, the moment makes a curvature, the curvature increases the bow. That loop converges to

$$M = \frac{N e_0}{1 - N/N_{cr} - N/S_v}$$

and the shear is its gradient: $V = \pi M / L$ for a half-sine bow.

Read the denominator, because it is the essay. It contains two flexibilities. $N/N_{cr}$ is the ordinary buckling amplification every slender member has. $N/S_v$ is the shear flexibility, and it is the lacing’s own contribution — so the amplification of the force the lacing must carry contains a term supplied by the lacing.

At 2,000 kN on this column: $N/N_{cr} = 0.522$, $N/S_v = 0.090$, the denominator is 0.388, the moment is 247 kN·m and the shear is 32.4 kN.

The common rule of thumb is $V = N/100$, which gives 20 kN. The rule understates by 62 per cent, and it understates by more the harder the column is working: at 1,500 kN the computed shear is 17.4 kN against the rule’s 15, a discrepancy of 16 per cent.

Which free body produced the shear #

The moment expression above is a formula, and the shear is read off it as a gradient, so it is worth doing the free body once — because the answer is where the diagonals’ force actually comes from and it is not obvious from the algebra.

Cut the column at a height $z$ and take the piece above. Crossing the cut is the axial force $N$, a moment $M(z)$, and a shear $V(z)$. The deflected shape is $y = a \sin(\pi z / L)$, so the moment at the cut is $N y$ — the axial load on its own lever arm, and nothing else.

The shear is then $\mathrm{d}M/\mathrm{d}z = N \, \mathrm{d}y/\mathrm{d}z$, which is the axial force times the slope of the column. That is the whole of it: a leaning column carries a transverse force equal to its axial load times how far it is leaning, in exactly the way a leaning column in a frame does.

The slope is largest at the ends and zero at mid-height, so the shear in the lacing is largest at the top and bottom of the column and zero in the middle — which is the reverse of where the moment is largest, and the reverse of where an intuition trained on beams puts it.

Two things follow that a uniform lacing hides.

The lacing at the ends is doing the work. A column laced uniformly is over-provided at mid-height and exactly provided at the ends, and any local weakness — a splice, a missed diagonal, a connection with slack — matters in proportion to how near an end it is.

And the end connections carry that shear out. The 32.4 kN has to be delivered into the base plate and the head detail, in a region where the lacing usually stops and a solid gusset takes over. The transition is a detail nobody computes, and it is where the shear is at its maximum.

Weak lacing makes the force it carries larger #

Now run the shear calculation on that column at 1,500 kN. $N/N_{cr}$ is 0.577, $N/S_v$ is 0.253 — the shear term has nearly tripled — and the denominator is 0.171. The moment is 421 kN·m and the shear is 55.2 kN, against 17.4 on the stiffer lacing.

The diagonals are a quarter of the area and are carrying three times the force. The stress in them has gone up by a factor of twelve.

At 2,000 kN the calculation returns a negative denominator, which is the arithmetic saying the column has passed its critical load and there is no equilibrium position to amplify towards. That is the correct answer and it is not a usable one: a column whose lacing has been trimmed is not carrying a large shear, it is not standing up.

The feedback is what makes this different from an ordinary amplification. Reducing the diagonal area reduces $S_v$, which reduces $N_{cr}$ and raises $N/S_v$ directly — two of the three terms in the denominator move the wrong way at once — and the resulting shear lands on the member that was reduced.

The angle is charged three times #

Three separate penalties arrive with that change and only the first is visible in the figure.

The critical load falls, from 3,832 kN to 3,456, because a flatter diagonal is a worse shear member — its axial extension turns into less transverse movement.

The design shear rises, because the denominator has got smaller: at 2,000 kN it is 0.275 rather than 0.388, and the shear is 45.8 kN against 32.4. That is 2.29 times the rule of thumb.

And the diagonal itself gets longer — 1,600 mm against 1,131 — so the member carrying that larger force is buckling over a length 41 per cent greater. A 300 mm² diagonal in compression at 1,600 mm is a slender member in its own right, and it is the one place in a laced column where a local check can quietly govern.

So 45° is not a convention. It is close to the maximum of a shallow curve: shallower loses shear stiffness fast, steeper gains almost none and costs material linearly. The received rule and the arithmetic agree, which is worth knowing because the received rule is usually stated as a detailing preference.

Two flexibilities, and which one is bigger #

The denominator has three terms and the middle one is doing most of the work, which is worth quantifying because it decides whether any of this matters on a given column.

At 2,000 kN on the hero column, $N/N_{cr} = 0.522$ and $N/S_v = 0.090$. The bending flexibility is nearly six times the shear one, so the amplification is mostly the ordinary buckling amplification that a solid column of the same critical load would have, and the lacing’s own contribution is a 9 per cent correction on top.

Trim the diagonals to 80 mm² and the two become 0.577 and 0.253 — the shear term has grown to 44 per cent of the bending one, and the correction has stopped being a correction.

This is the same series-flexibility argument the rung below this one makes about the critical load, arriving one level up. There the two flexibilities added to give $1/N_{cr} = 1/N_E + 1/S_v$; here they add again inside an amplification, and the second appearance is the one that decides a member size.

A quantity that is a small correction to a stiffness can be a large correction to a force, because the force is computed through a difference rather than a sum: $1 - 0.522 - 0.090$ is 0.388, and removing the last term would give 0.478 — a 23 per cent change in the answer from a term that is 9 per cent of a denominator that started at 1.

That sensitivity is the general reason second-order calculations are unforgiving. Every term in the denominator is subtracted from one, so the closer the column is to its capacity the more a small flexibility is worth, and the amplification’s sensitivity to any of its inputs rises without limit as the denominator approaches zero. It is the same arithmetic that makes a slender member’s deflection unbounded rather than merely large.

The chord between two lacing points #

The panel length is set by the lacing angle, and it is also the buckling length of the chord as an individual member.

The chords here have $i = 28.3$ mm. At 45° the panel is 800 mm and the chord’s local slenderness is 28.3; at 30° the panel is 1,386 mm and the slenderness is 49; at 60° it is 16.3. The whole column’s effective slenderness is 66 to 69 across the same range.

So the local mode does not govern on this column, at any of these angles, and that is the usual state of affairs — which is why the check is often omitted.

Where it stops being usual is where the panel is set by something other than the lacing.

That figure is the reason a shear stiffness matters more than a local check: the batten arrangement is catastrophically worse globally while its local slenderness is unremarkable. A member can be perfectly satisfactory piece by piece and be a tenth as strong as its second moment of area says, and the piece-by-piece check is the one an intuition trained on solid sections reaches for.

What the rule of thumb is actually for #

The $N/100$ rule is old, it is in several codes as a floor rather than as the calculation, and it is worth being fair to it.

It is a shear stiffness requirement written as a strength one. A lattice sized to carry one per cent of the axial load, at a sensible angle, with diagonals stocky enough not to buckle, will have a shear stiffness large enough that $N/S_v$ is small — and if $N/S_v$ is small then the amplification is the ordinary one and the rule’s own answer is close to right.

It is self-consistent when it is satisfied and wrong when it is not, which is the property of a rule that encodes a proportion rather than a mechanism. It fails in exactly the cases this essay has drawn: a lacing trimmed for weight, a shallow angle chosen for fabrication, a column loaded near its capacity.

The honest procedure is the loop the rule replaces. Assume a lacing, compute $S_v$, compute $N_{cr}$, compute the amplified moment and its shear, check the diagonal for that shear including its own buckling over the panel diagonal, and go round again if the diagonal changed. Two iterations is usually enough, and the reason it converges quickly is the same reason the rule usually works: $N/S_v$ is a small term whenever the lacing is not silly.

The same loop, three places on this site #

The shape of this argument — a stiffening element whose own flexibility amplifies the force it must carry — is not confined to laced columns, and recognising it elsewhere is most of what makes it worth writing down.

A bracing member restraining a column is the closest relative. The brace is sized for a force that is the restrained load times an angle, the angle depends on how far the column has already moved, and how far it has moved depends on the brace’s stiffness. A brace that is strong enough and too flexible does not restrain, and the force it is asked for grows as its stiffness falls — which is why bracing rules specify a stiffness as well as a strength, and why the strength requirement alone is never sufficient.

A stiffener on a plate is the same thing across a surface: the rib is asked to hold a line straight, the force it carries depends on the out-of-flatness that has developed, and a rib too flexible to hold the line is loaded harder than one that holds it.

And a frame’s own second-order sway is the version at building scale. The lateral force on a bracing system includes a term proportional to the vertical load times the sway, the sway depends on the bracing system’s stiffness, and the stiffest path takes the load in a way that feeds back on itself.

In all three the practical instruction is identical and slightly counterintuitive: when a stiffening element is found to be overloaded, the fix is to make it stiffer rather than stronger. Adding area to a diagonal reduces the force in it. Adding yield strength does not.

That is a rare property. Most members in most structures carry a force decided elsewhere, and making them stronger is the whole of the repair. These carry a force decided by themselves.

What to carry away #

The shear is an imperfection effect. A straight column has none, and the design value comes from an assumed bow of $L/500$ that is a fabrication tolerance rather than a load.

The lacing amplifies the force it carries. Its own flexibility appears in the denominator of the amplification, so trimming it raises the demand on itself — three times the shear on a quarter of the area, in the case drawn.

The angle is charged three times over. Shallower lacing loses critical load, gains design shear, and lengthens the member that has to carry it.

And the rule of thumb is a stiffness rule in strength clothing. It is right when the lacing is generous and understates by 62 per cent when it is not.

Where the model stops #

The bow is a single half-sine of assumed amplitude. A real column’s imperfection is whatever the fabrication produced, in a shape nobody measured, and the calculation is a convention that has been calibrated against tests rather than a description of one column.

Nothing here is inelastic. Every critical load on this page is an elastic one, and a stocky built-up column reaches its squash load first — which is the same competition a solid column has with a second flexibility added to one side of it.

The connections are rigid and free of slip. A bolted lacing slips into bearing before it carries anything — the hole goes oval and the bolt finds its far side — and that slip is a shear flexibility in series with the elastic one, larger on a real column than the difference between single and double lacing. It is also the reason lacing connections are so often specified as preloaded when nothing else on the member is.

The bow is assumed and not measured. A column built to the wrong length or out of straight carries the imperfection it actually has, and $L/500$ is a tolerance the fabricator is asked to stay inside rather than a value anybody checks on the finished member. Every shear on this page is proportional to that number, so the design shear is proportional to a workmanship limit.

And the stiffness is taken as uniform along the column. An average stiffness is not a safe stiffness for a member whose lacing changes — a splice region, a heavier panel at the base — and the shear stiffness that belongs in the amplification is the one at the ends, where the shear is largest, rather than the mean.

And the two planes are treated as independent. A four-chord tower laces about two axes at once, its lacing planes share chords, and the shear stiffness in one direction is not independent of the deformation in the other.

The ladder from here #

Later rungs on this anchor: Haringx against Engesser, and the elastomeric bearing where the two formulations disagree most. Bolt slip as a shear flexibility, and the built-up column whose lacing is nominally adequate and whose connections are not. Four-chord masts and towers, where the lacing planes interact. Built-up columns in timber, where the connection is nails and the slip modulus does the whole of the lacing area’s job. And the general case this belongs to — any member whose stiffness is assembled from a bending term and a shear term in series, which includes a shear wall, a Vierendeel frame and a sandwich panel.

Engesser published the shear correction in 1891, after built-up columns had been failing for two decades at loads their second moment of area said were safe. What took longer to write down was the design shear — the number the lacing is sized for — because it is not a load at all. It is the consequence of an imperfection, computed through a stiffness the calculation is trying to determine, and for most of the twentieth century the profession used a fraction of the axial load instead and got away with it because the fraction was chosen generously.