The torsion that goes away if you let it

A floor beam frames into the side of an edge beam. The floor beam wants to rotate at that end; the edge beam resists it by twisting; and a frame analysis duly reports a torque in the edge beam, which somebody then has to design for.

The number that comes back is entirely a property of the stiffnesses that were typed in. Halve the edge beam’s torsional rigidity and the torque halves. Set it to nothing and the torque disappears, along with any obligation to reinforce for it — and the structure still stands, because the floor beam simply spans a little further before it finds a support.

That is not true of every torque, and telling the two apart is the whole of this subject.

Two kinds, and only one of them is optional #

Equilibrium torsion is torsion that exists because there is no other load path. A canopy cantilevering off the side of a beam applies a torque per metre of $w a^2/2$; the beam carries it to the columns and the columns carry it to the ground, and there is no arrangement of stiffnesses that changes the number. Take the torsional resistance away and the canopy falls.

Compatibility torsion is torsion that exists because two members meeting at a joint are obliged to rotate together, and one of them happens to be stiff in twist. It is a consequence of stiffness, not of load, and the load it corresponds to has somewhere else to go.

The distinction is old and its practical content is a permission: a compatibility torque may be reduced, or ignored entirely, provided the members that pick up what it sheds are checked for having picked it up. An equilibrium torque may not be reduced by anything.

Which free body produced the number #

Take the joint as the free body: a short length of the edge beam with the floor beam framing into it.

Three things act on it. The floor beam delivers a moment about the edge beam’s axis. The edge beam to the left resists by twisting; the edge beam to the right does the same. The joint has one rotation, shared by all three, and the moment splits between the branches in proportion to their stiffnesses.

The edge beam, held against twist at both columns and loaded at midspan, offers

$$K_t = \frac{4GJ}{L}$$

— two halves of length $L/2$, each of stiffness $2GJ/L$, in parallel. The floor beam, pinned at its far end, offers $K_b = 3EI_b/L_b$. And the moment they are sharing is the floor beam’s fixed-end moment, $wL_b^2/8$.

For the members drawn: a 400 by 700 edge beam spanning 8 m has $GJ$ of about 135,000 kNm², so $K_t = 67{,}500$ kNm per radian. A 300 by 600 floor beam spanning 7.5 m has $EI$ of about 162,000 kNm², so $K_b = 64{,}800$. The fixed-end moment at 28 kN/m is 197 kNm, and the edge beam takes

$$T = 197 \times \frac{67{,}500}{67{,}500 + 64{,}800} = 100\;\text{kNm}$$

Slightly more than half, from an arrangement nobody would look at twice.

That is the same arithmetic as moments passed around a joint until they balance, each branch taking its share by stiffness. A torsion member is one more branch, and the only thing unusual about it is that its stiffness is $GJ$ rather than $EI$.

What cracking does to $GJ$, and why it is worse than what it does to $EI$ #

The 135,000 kNm² above is the uncracked value, and reinforced concrete does not stay uncracked in torsion for long. Once a diagonal crack has formed, a solid section stops behaving as a solid section: what is left is a thin-walled tube of concrete with the reinforcement in it, and its torsional stiffness is a small fraction of what it was.

The fraction is far smaller than the corresponding one in bending. A cracked section in bending keeps something like a third to a half of its flexural stiffness, because the compression zone is intact and the reinforcement replaces the tension zone. A cracked section in torsion keeps perhaps a fifth to a quarter, because the mechanism it used to resist by — shear stress circulating through the whole solid area — is not available to a cracked one at all.

The reason the two differ is worth stating rather than quoting. A cracked section in bending still has a compression zone doing exactly what it did before, with steel taking over the tension; nothing analogous happens in torsion, where the resisting mechanism was the whole section circulating shear and there is no reduced version of it to fall back on.

At a quarter of the uncracked stiffness the edge beam offers 16,900 kNm per radian instead of 67,500, and its share falls from 51% to 21%. The torque falls from 100 kNm to 41 — a factor of 2.5 from a factor of 4 in stiffness, because a distribution factor is not linear in the thing being distributed.

Where the shed torque goes #

It does not evaporate. The floor beam’s end moment was the thing being shared, and whatever the edge beam declines to take, the floor beam keeps.

Write it as an identity rather than as a result. The floor beam’s bending moment is the free parabola with a straight line subtracted, so at midspan

$$M_{mid} = \frac{wL_b^2}{8} - \frac{T}{2}$$

and the two always add back to 197 kNm whatever $T$ turns out to be. At $T = 100$ the midspan moment is 147; at $T = 41$ it is 177. Twenty per cent more, on a beam that was designed for the first number.

This is what makes ignoring compatibility torsion safe rather than merely convenient. The permission is conditional on checking the member that inherits the load, and the check is a bending check on a beam whose end restraint has been reduced — which is arithmetic anybody can do.

The general form of the trade is the same everywhere in the subject. A support moment and a span moment are one free moment cut at different heights, so anything that reduces one raises the other by half as much, and every redistribution argument in this collection is that identity with a different reason for moving the line.

The variable is a ratio, and either half of it will do #

The share the edge beam takes is $K_t/(K_t + K_b)$, which means the designer has two levers and they are equally effective. Making the edge beam softer in torsion reduces the torque; making the floor beam stiffer in bending reduces it just as much, and does so without touching the member that was in trouble.

That second lever is worth naming because it is invisible from inside the member being designed. Deepening the floor beam from 600 to 750 mm multiplies its $EI$ by $1.25^3$, to 316,400 kNm², and the deeper beam wins more of the argument at the joint — taking the torque off its neighbour by taking the moment itself. Nothing about the edge beam changed. It is the stiffest path takes the load again, and the third power in a bending stiffness is why a modest change of depth is a large change of share.

The third lever is the one that removes the argument rather than winning it. Detail the floor beam’s connection to the edge beam as a genuine pin and $K_b$ becomes zero: the floor beam has no end moment to share, the edge beam attracts nothing, and there is no torsion to design for at all. Whether that is available is a question about the connection rather than about either member.

The rotation is the price #

The edge beam sheds torque by rotating more, and the rotation is the quantity nobody sees in the analysis output.

At the uncracked stiffness the joint turns through $T/K_t = 100/67{,}500 = 1.49$ milliradians. At a quarter of that stiffness it turns through 2.41 — more rotation, less torque, which is the signature of a compatibility effect and is exactly backwards from what an equilibrium torque does.

A milliradian and a half over a 7.5 m floor beam is 11 mm of extra deflection at the far end of it, which is not the kind of number that fails anything. Over a much longer floor beam, or with an edge beam that has genuinely gone soft, it is a crack in a partition and a door that catches. The permission to ignore the torque is a strength permission, not a serviceability one.

What an edge beam offers a floor beam is a joint with a stiffness rather than a joint that is rigid or pinned, and the classification axis on that argument is the same axis as this one: a connection stiff enough to count as fixed, soft enough to count as pinned, or somewhere in the middle where the answer depends on a number nobody measured.

Where along the spandrel the beam frames in #

The 100 kNm above is for a floor beam framing in at midspan, and a real floor has beams at three-metre centres along the whole edge. They do not all get the same answer, and the variation is larger than any of the levers just described.

The torsional stiffness a spandrel offers at a station is two lengths in parallel:

$$K_t = \frac{GJ}{a} + \frac{GJ}{L-a} = \frac{GJ\,L}{a(L-a)}$$

which at midspan is the $4GJ/L$ used above and rises steeply toward either column. At the quarter point it is $16GJ/3L$, a third stiffer; a metre from an 8 m column it is $9.14\,GJ/L$, 2.3 times the midspan value.

Feed those through the same distribution:

A floor beam near a column delivers 39% more torque than an identical one at midspan, and keeps 39% less of its own end moment. So a row of nominally identical floor beams are not identical: the ones near the columns are appreciably fixed at their spandrel ends and the ones in the middle are close to simply supported, and the midspan moments along one edge of a floor vary by a fifth with nothing changing but which column each beam is near.

The spandrel’s own diagram is the other half of it. Each floor beam delivers its torque at a point, so the spandrel’s torque diagram is a staircase rising from midspan toward each column, and the value at the column is the sum of every share on that side. Five beams at 3 m centres on an 8 m spandrel is not a realistic count, but the shape is the point: the torsion to be reinforced for is an accumulation, and it is largest exactly where the shear and the hogging moment are largest too.

The steel case inverts the permission #

Everything above depends on cracking, which is how a concrete spandrel sheds what it declines to carry. Steel does not crack, so the stiffness that attracted the torque is the stiffness that keeps it — and the compatibility torque in a steel spandrel is not negotiable in the way a concrete one is.

That sounds alarming and is usually nothing, because of the two orders of magnitude an open section is down. A 457 deep universal beam has a torsion constant of about $5.2\times10^5$ mm⁴, so $GJ \approx 42$ kNm² against the concrete spandrel’s 135,000, and $K_t = 21$ kNm per radian against a floor beam’s 64,800. Its share of the joint moment is three hundredths of one per cent. The steel spandrel does not shed the torque; it never attracts any.

The case that does need designing is the one chosen for its torsional strength. A 400 × 200 × 12.5 rectangular hollow section has $J \approx 2.3\times10^8$ mm⁴ and $GJ \approx 18{,}600$ kNm², giving $K_t = 9{,}300$ and a share of 13% — about 25 kNm, permanently, with no mechanism available to reduce it.

Which is an inversion worth carrying. In concrete, torsional stiffness is something a member has and then loses under load, so the permission to ignore compatibility torsion is the physics rather than a concession. In steel, specifying the section that is good at torsion is what creates the obligation to design for it — and a designer who reaches for a box because the member “has some torsion in it” has increased the torsion by reaching.

The equilibrium case, and why it will not negotiate #

A canopy 2.2 m deep carrying 12 kN/m² hangs off the same edge beam. Each metre of the beam receives a torque of

$$t = \frac{w a^2}{2} = \frac{12 \times 2.2^2}{2} = 29\;\text{kNm per metre}$$

and over an 8 m span with columns at each end, the torque delivered to each column is $tL/2 = 116$ kNm.

There is no stiffness in that expression. The beam could be steel or concrete, open or closed, cracked or uncracked, and 116 kNm would still arrive at the column, because the sums have to cancel on a free body that contains the canopy and one column and nothing else.

One pair of free bodies settles it. Cut the canopy off and the piece that is left has a moment on its cut face that nothing but the beam can supply. Cut the floor beam off in the compatibility case and the piece that is left is in equilibrium already, with no moment on the cut face at all. The free body is a choice, and that pair of choices is what the whole distinction reduces to.

The two produce identical output from an analysis. Both appear as a torsional moment in a member; both have units of kNm; both sit in the same column of the same table. The difference is not in the result, it is in what the result depends on, and the only way to know is to ask what happens to the structure if the torsional stiffness is set to zero. If the answer is the load takes another route, it is compatibility torsion. If the answer is the structure falls down, it is not.

What the pictures cannot show #

The stiffness reduction is drawn as a single number multiplying $GJ$, and cracking is not a number. It starts somewhere along the member, spreads as the load grows, and leaves the section with a different stiffness at every station. What is drawn as a point on a curve is really a beam whose torsional stiffness varies along its length and with its history.

Nor can the figures show when it happens. A spandrel that has never been loaded to its cracking torque is at the uncracked value, so a serviceability calculation and an ultimate one are entitled to different answers from the same member — and the torsion that matters for cracking a partition is the larger one, at the stage where it has not yet been shed.

The rotation is drawn as a number at a point, and a real spandrel twists progressively along its length: the 1.49 milliradians quoted is the relative rotation between the joint and the columns, and the shape of the twist between them is the thing the figure of the torque path shows and this one does not.

And the joint is drawn as a point. The floor beam actually frames into the side of the edge beam over a real width, the torque is delivered as a couple of shear forces on two faces, and the region around it obeys none of the assumptions any of this rests on — which is the standing exclusion every member calculation on this site inherits.

The assumption the figures rest on #

Every stiffness here is elastic and constant along the member. That is the assumption that lets one distribution factor stand for a whole joint, and it is doing more work than usual: a torsion member’s stiffness falls further and faster on the way to its ultimate state than any other branch in a frame, so the distribution factor computed at working load is not the one that governs at collapse.

The second assumption is that the far end of the floor beam is pinned, which fixes $K_b$ at $3EI/L$ and the fixed-end moment at $wL^2/8$. Make it continuous and both change — $4EI/L$ and $wL^2/12$ — and the edge beam’s share moves with them.

The competing stiffness matters as much as the torsional one, and a stiffer floor beam is a way of reducing the torque that has nothing to do with the member carrying it.

The minimum that is not a calculation #

If a compatibility torque may be ignored, the member still cracks — it just cracks and then stops attracting the load. What stops the crack from becoming a failure is reinforcement that was never designed for a computed force: closed links and longitudinal bars at the corners, provided as a minimum rather than as an answer.

That is an unusual thing to find in this subject, and it is worth being honest about what it is. It is not a strength calculation whose number happens to be small. It is a decision to let a member crack, on the condition that when it does, it holds together well enough to hand its load to something else. The reinforcement is there to make the redistribution survivable rather than to resist anything.

The ladder from here #

Later rungs on this anchor: the torsion member as a branch in a moment distribution done by hand, where the torsional stiffness enters the same table as every other. The interaction of torsion with shear in a cracked section, where the two demands on the same links add rather than compete. Warping torsion in the compatibility case, where a spandrel restrained against warping at the columns is very much stiffer than $4GJ/L$ says and attracts correspondingly more. The threshold question — how large a compatibility torque has to be before shedding it is not the right answer. And the case that sits between the two kinds, where a torque is equilibrium torsion for one load case and compatibility torsion for another on the same member.