A deck is a spring, not a wall

The movement with no limit against it computed the twist of an open-section beam loaded off its shear centre and found it large: an 8 m beam carrying 12 kN/m at 75 mm from its shear centre moves its flange tip sideways by more than half its own vertical sag. It then offered the usual reassurance, that “a deck screwed to the top flange makes the twist calculation on this page irrelevant”, because the finished beam has continuous restraint along its length and cannot twist at all.

That reassurance is a statement about stiffness, and it can be tested. A deck fastened to the top flange resists the flange’s rotation — when the beam twists, the deck has to bend and its fasteners have to give — and it does so with some moment per metre of span for each radian of twist. That is a distributed rotational spring, and whether it makes the twist irrelevant depends on how stiff it is against the beam’s own resistance to twisting. The comparison turns out to be close, and on ordinary sheeting the deck is nearer the beam’s stiffness than it is to a wall.

The deck as a bed of rotational springs #

An open section resists a distributed torque $m$ in two ways, by shear circulating round its walls, the $GJ$ term, and by its flanges bending apart, the warping term $EI_w$. The first is feeble in an open section — the slit that costs a factor of six hundred is the reason — and the second exists because the section cannot stay flat when it twists, and resists being made to warp unevenly along its length. A deck resisting rotation adds a third term, proportional to the twist itself:

$$EI_w\,\theta'''' - GJ\,\theta'' + k_\theta\,\theta = m.$$

This is the beam that sits on the ground turned about its own axis: the beam’s deflection has become its twist, its bending stiffness has become its two torsional stiffnesses, and the ground’s modulus has become the deck’s rotational restraint $k_\theta$, in kNm per metre of span per radian. The solution is the same kind of function, a load shared between a member and its bed, with the member carrying the torque near its supports and the bed carrying it in the middle.

The beam here sits on forks — twist held at the ends, warping free — which is what a fin plate or a pair of web cleats provides. With nothing fastened to it, it twists 6.25° at mid-span and its flange tip moves 21.8 mm sideways. A deck of 5 kNm per metre per radian brings that to 4.20°. That is a real improvement, a third of the twist removed, and it is nowhere near “irrelevant”.

The number to compare the deck with #

The deck’s stiffness only means something against the beam’s own, and the beam’s own has a clean expression. A fork-ended beam twisting in its first half-sine resists a distributed torque of that shape with a stiffness, per metre, of

$$k_1 = GJ\left(\frac{\pi}{L}\right)^2 + EI_w\left(\frac{\pi}{L}\right)^4,$$

which for this 8 m section is 7.0 kNm per metre per radian from Saint-Venant torsion and 3.4 from warping, 10.4 in all. A deck of $k_\theta$ adds in parallel, so the twist falls roughly as

$$\frac{\theta}{\theta_0} \approx \frac{1}{1 + k_\theta/k_1}.$$

The solved curve and the one-mode estimate barely differ, which is the useful fact: the twist halves when the deck is as stiff as the beam, and the beam’s stiffness is a two-term formula from a section table. The restraint that halves this beam’s twist, found by solving the full equation, is 10.2 kNm per metre per radian; the estimate says 10.4.

What a real deck supplies is the other half of the comparison, and it is not a property of the deck alone. Its rotational restraint is two springs in series, the sheeting bending between beams and the fasteners giving at the flange, and two flexibilities in series make the softer one govern. The fasteners are nearly always the softer. The European code for cold-formed steel tabulates their stiffness for trapezoidal sheeting screwed to a flange at a few kNm per metre per radian, depending on the spacing and the washers. That is the same order as this beam’s own 10.4, which is why the deck removes a third of the twist rather than all of it.

The 8 m beam by hand #

The whole comparison fits in four lines. For the section drawn, $GJ = 81{,}000 \times 5.6\times10^5 = 4.54\times10^{10}$ N·mm², and $(\pi/8{,}000)^2 = 1.54\times10^{-7}$ per mm², so the Saint-Venant part of $k_1$ is 7,000 N·mm per mm per radian, 7.0 kNm/m/rad. The warping part is $205{,}000 \times 7.0\times10^{11} \times (\pi/8{,}000)^4 = 3{,}420$, another 3.4. A deck of 5 then gives $1/(1 + 5/10.4) = 0.675$ of the free twist, against 0.673 solved: 4.2° at mid-span and a flange tip moving 14.7 mm instead of 21.8.

The same arithmetic answers the design question in reverse. To hold the free twist $\theta_0$ down to an allowed $\theta_a$ takes a deck of about $k_1(\theta_0/\theta_a - 1)$, a requirement that grows without limit as the allowance tightens, because the last part of any twist is the hardest to remove.

A beam held against warping is helped less #

End restraint and a deck compete for the same job. A beam whose ends are held against warping is already much stiffer in torsion — 2.55° rather than 6.25° under the same load — and its stiffness is concentrated in the shape the deck would otherwise resist. The same deck of 5 kNm per metre per radian removes 16 per cent of its twist, and halving it takes 25.4, two and a half times the fork-ended figure.

So the two ways of stiffening a beam against twist do not add as a designer might hope. A detail that welds an end plate across both flanges and a deck screwed along the top flange each remove a share of the twist, and the second removes less once the first is in place. Which to rely on is a question of which is actually there: the end plate is on the drawings and is there on the day the steel is erected, while the deck’s contribution depends on fasteners that are installed later and have a stiffness nobody measures.

Where the deck carries the torque #

The deck resists torque in proportion to the twist, and the twist is zero at the supports, so the deck carries its share in the middle of the span and the beam carries the rest outward to its ends. That makes the division of labour a question of position as well as of stiffness. With a deck of 5 kNm per metre per radian, the deck takes 26 per cent of the whole span’s torque but about 40 per cent of it at mid-span; near the supports the beam takes all of it.

That second fact matters to the fixings. A deck restraining a beam is transmitting a moment through its fasteners into the flange, and at mid-span with a deck of 20 kNm per metre per radian that moment is 734 N·mm per mm of span, 0.73 kNm per metre. The restraint that is free for the movement check is a strength demand on the screws, and it is largest exactly where the flange is also carrying its peak bending stress. A deck credited with holding a beam straight has to be fixed as a structural connection rather than as cladding.

Put in terms of a single screw, the demand is not small. With a deck of 5 kNm per metre per radian the mid-span moment is 367 N·mm per mm; fasteners at 300 mm centres each transmit about 0.11 kNm, and if the sheet pivots about the edge of a 178 mm flange with the screw on the centreline, that is a pull of about 1.2 kN in each screw. It is the same order as the pull-out and pull-through resistances such screws are selected for under wind suction — so a deck asked to restrain twist is asking its fixings to do a second job at a load comparable to the first.

The supports see the complement. The torque that reaches each end is the beam’s share, half of $m L$ less what the deck took, so a stiffer deck relieves the end connections too: with a deck of 20 each end carries about 46 per cent of what it would carry with none. That is a free benefit, as long as the fixings deliver what the calculation assumed.

The same deck does more on a longer beam #

The beam’s first-mode stiffness has $L$ in its denominator twice over: $GJ(\pi/L)^2$ falls as the square of the span, and $EI_w(\pi/L)^4$ as the fourth power. The deck’s restraint is per metre of span and does not depend on the span at all. So a longer beam is softer in torsion by at least the square of its length, and the same deck becomes stiffer relative to it at the same rate.

The figure shows what that does. On a 4 m beam, whose own stiffness is 82 kNm per metre per radian, a deck of 5 removes 6 per cent of the twist, which is to say nothing a designer would notice. On a 16 m beam, whose own stiffness is below 2, the same deck removes 73 per cent, and the flange tip’s movement falls from 115 mm to 31. The restraint that halves the twist falls from 81.7 kNm per metre per radian at 4 m to 1.9 at 16 m. A long, torsionally weak beam is exactly the one a modest deck can hold, and a short, stiff one is the one it cannot help.

This is the pattern of continuous restraint against buckling, where an elastic bed along a strut matters more the longer the strut is, until the strut’s own length drops out of the answer and the bed alone sets the wavelength. The same thing happens here: on a long enough beam the twist in the middle of the span is simply $m/k_\theta$, the torque over the deck’s stiffness, and the beam’s section properties no longer enter. A 16 m beam on a deck of 5 kNm per metre per radian is already more than halfway there.

The restraint a limit asks for #

Since no code gives a twist limit, the only comparison available is the one the first essay on this used: the flange tip’s sideways movement against the limits that apply to the vertical sag. Borrowed that way, the question becomes a requirement on the deck. On the 12 m beam, whose flange tip moves 59.7 mm with nothing fastened to it, holding it to span/360 needs a deck of 2.9 kNm per metre per radian, and to span/500 one of 5.5.

Those are ordinary numbers, within what screwed sheeting supplies, and that is the second useful fact: on a long open beam the deck is not a refinement but the thing that makes the beam serviceable at all, and it does so with fasteners that were specified for wind uplift rather than for torsion. The hand rule gives the same numbers: the 12 m beam’s own stiffness is $k_1 = 3.8$ kNm per metre per radian, and $3.8 \times (59.7/33.3 - 1) = 3.0$ and $3.8 \times (59.7/24.0 - 1) = 5.6$, against 2.9 and 5.5 solved. Double the eccentricity to 150 mm and the requirements become 9.4 and 14.2 kNm per metre per radian, at or beyond what screwed sheeting reliably provides. The curve flattens as the deck stiffens, so each tightening of the limit costs more restraint than the one before.

Which free body produced the number #

Cut a short length of the beam, $dx$, at mid-span. Acting on it are the torque applied by the eccentric load, $m\,dx$; the difference between the internal torques on its two faces, which is what the beam carries toward its supports; and the moment the deck applies through the fasteners, $k_\theta\,\theta\,dx$, resisting the rotation. Equilibrium of that slice about the beam’s axis is the whole of the governing equation. What the deck contributes is a local restoring moment proportional to the local twist, which is why it cannot act at a support where nothing twists and why it acts most where the beam twists most.

Two limits of that slice are worth knowing by hand, because they bracket everything the figures show. With no deck the slice is carried entirely by the beam, and the mid-span twist has closed forms: $(m/GJ)[L^2/8 - (1 - \operatorname{sech}(kL/2))/k^2]$ on forks, 6.25° here, and $(m/GJ)[L^2/8 - (L/2k)\tanh(kL/4)]$ with the ends held against warping, 2.55°, where $k = \sqrt{GJ/EI_w}$. With a rigid deck the slice is carried entirely by the deck, and the twist is $m/k_\theta$, independent of the section. Every curve on this page runs between those two, and which end it sits nearer is decided by the ratio $k_\theta/k_1$ alone.

Where the model stops #

The deck is a uniform, two-sided spring. Real sheeting is fastened at discrete troughs, so its restraint is a row of point springs that approximate a continuous one only when their spacing is short against the length over which the twist varies — several metres here, so the approximation is good. It is less good in the other respect: sheeting resists the flange rotating one way much better than the other when the load pulls the flange away from the sheet, and a fastener whose washer lifts off is no spring at all.

The deck is linear. Fastener stiffness falls as the connection deforms, and it falls further under repeated loading, so a deck credited with 5 kNm per metre per radian on the day it is fixed may supply less after years of thermal cycling. The requirement read off the last figure is for the stiffness the deck has when the movement matters, not when it was installed.

Lateral movement is left out. The deck also holds the top flange against moving sideways, which moves the effective centre of twist from the shear centre to the flange and changes the problem in a way this model does not include. For a load applied on the top flange that restraint helps further; for one hung from the bottom flange it can make the bottom flange’s movement the larger.

And the construction stage is not the finished one. The deck earns its restraint only once it is fixed. Laid loose on the beam, it restrains nothing, and the beam twists by the whole of its no-deck value under whatever is standing on it — a pump, a bundle of decking, a concrete pour on one side. That is the case where the calculation matters most and where the reassurance does not apply at all.

There is one escape the deck does not provide and the load sometimes does. Where the torque exists only because the beam is stiffer than what it supports — a slab bearing on one side that could rotate a little instead — letting the beam twist relaxes the torque, and the torsion goes away if it is allowed to. A deck resisting the twist works against that relaxation. For a torque fixed by statics, such as a facade hung on brackets, the deck is the only help available; for one that exists by compatibility, stiffening the restraint attracts more of it.

What the pictures cannot show #

That the deck’s stiffness is a number nobody measures. The fastener stiffness in the European code is tabulated from tests on particular sheets, washers and flange thicknesses, and a site that substitutes a different screw, omits every second fastener or fixes to a painted flange has a restraint the figures do not describe. A beam that relies on its deck to meet a movement limit relies on a detail that is inspected for weathertightness and never for stiffness.

Nor can they show the effect on buckling. A deck that restrains the top flange’s rotation also raises the load at which the beam buckles laterally, and codes credit it for that with the same kind of rotational stiffness. The two uses of the one spring are not independent: a deck that is taking a torque from an eccentric load has used some of its fasteners’ capacity before any buckling restraint is asked of it.

Still open: the deck that meets the beam at a point #

Every figure here spread the deck’s restraint evenly along the span. A precast plank bearing on one side, a deck spanning onto the beam from only one direction, or a secondary beam framing in at mid-span each restrain the twist over a length much shorter than the span, or at a point. Whether a single rotational restraint at mid-span, as stiff in total as the distributed deck, holds the beam as well — or whether a distributed restraint is worth several times its total stiffness concentrated at one point, as a stiffener’s rigidity is worth more in some positions than in others — is the question the uniform deck leaves open.