Module 5 · Lesson 5.4

Torsion

The same truss, wrapped round a tube — and the one question you must answer before designing for it at all.

Why this matters

Torsion is the action most often designed for unnecessarily, and occasionally not designed for when it had to be. Both errors come from the same place: not asking whether the twist is holding the structure up or merely happening to it. Answer that first, and half of torsion design disappears.

By the end of this lesson you should be able to

Distinguish equilibrium torsion from compatibility torsion, and say why it matters
Explain why a solid section is analysed as a thin-walled tube
Apply Bredt's formula to find the shear flow
Design closed links and the longitudinal steel that must accompany them
Combine torsion with shear through the interaction check

What you should already know

The variable-angle truss (this module)
Torsion of thin-walled sections (Structural Analysis Fundamentals, Module 11)
Shear as diagonal tension (this module)

The question that comes first

Before any calculation, decide which kind of torsion you have.

Equilibrium torsion is required for the structure to stand up. A canopy cantilevering off the side of a beam has nowhere else to send its load: the beam must twist, and if it cannot resist the twist the canopy falls. There is no alternative load path, so it must be designed for in full, however small it is.

Compatibility torsion arises only because members are joined. A floor beam framing into an edge beam rotates its end as it deflects, and the edge beam twists because it is attached. Nothing depends on that twist.

The difference matters because of what happens when a member cracks in torsion. Torsional stiffness after cracking collapses to a small fraction of the uncracked value — far more dramatically than flexural stiffness does. So a member carrying compatibility torsion simply cracks, stops resisting the twist, and the rotation is accommodated elsewhere. The load finds another path because there was one all along.

Compatibility torsion may be ignored for strength, provided minimum reinforcement is present to keep the cracking controlled. Equilibrium torsion may not be ignored at all.

Mistaking equilibrium torsion for compatibility torsion removes a load path that has no alternative. That is the error that matters.

Bredt's formula

τ t = \frac{T _{E d}}{2 A _{k}}

What it calculates: The shear flow round a closed thin-walled tube under torsion.

$T_{E d}$: Design torsional moment (N·mm)
A_k: Area enclosed by the CENTRELINE of the wall (mm²)
t: Wall thickness (mm)
$τ$: Shear stress in the wall (N/mm²)

This assumes

The tube is closed — an open section behaves completely differently and far worse
Thin wall, so the stress is uniform through the thickness
PURE EQUILIBRIUM: no material property appears

In plain terms: The shear flow τt is CONSTANT round the tube, which is the whole content of the formula. Note that A_k is the area enclosed by the wall centreline, not the gross section area — using the gross area overstates A_k and understates the shear flow, always unconservatively.

The same truss, wrapped round

Once the shear flow is known, the design is the truss of the previous lessons applied to each wall of the tube. Two results follow, and the second is the one that gets forgotten.

Closed links carry the shear flow across the diagonal cracks:

A_sw/s = T_Ed / (2 A_k f_ywd cot θ)

The links must be closed, with a proper lap or hook. A torsional shear flow runs continuously round the tube, and an open link has nothing to carry the flow across its gap — an open link in torsion is close to useless.

Longitudinal steel carries the axial component of the inclined struts:

ΣA_sl f_yd / u_k = T_Ed cot θ / (2 A_k)

The struts spiral round the member, so they push along it as well as around it. Something has to carry that push, and it is longitudinal reinforcement distributed round the whole perimeter — including a bar in every corner, because that is where the strut forces are resolved. This steel is additional to the flexural reinforcement.

Take care

In torsion, flattening the strut is not free

In beam shear, flattening the strut towards cot θ = 2.5 was almost always right: it engaged more links and cost only some strut capacity.

In torsion the trade is different, because the strut angle appears in both reinforcement requirements and in opposite senses:

	cot θ = 1.0	cot θ = 2.5
Links, A_sw/s	0.442	0.177
Longitudinal, ΣA_sl	644 mm²	1609 mm²

Flattening the strut from 45° to 22° cuts the link steel by 60% and multiplies the longitudinal steel by two and a half. There is a genuine optimum, and it is not automatically the flattest angle. For a member also carrying flexure — where longitudinal steel is already present and may have spare capacity — a flatter angle can still win. For one in pure torsion it usually does not.

Worked example

An edge beam carrying a cantilever slab

Given

Edge beam 350 mm wide × 600 mm deep, C30/37
35 mm cover, H10 closed links, H20 longitudinal bars
Design torsion T_Ed = 45 kNm (equilibrium torsion — the slab cantilevers off it)
Design shear V_Ed = 180 kN at the same section

Find

Whether the section is adequate, and the reinforcement required.

Assumptions

cot θ = 2.5, the flattest permitted — chosen here so the comparison with shear is direct
The section is not hollow, so the equivalent tube is used

Predict first

A floor beam frames into an edge beam. The floor beam deflects, rotating its end, and twists the edge beam. What must the edge beam be designed for?

Practice

A 400 × 500 solid section has 40 mm cover, H10 links and H20 bars. What is the equivalent wall thickness t_ef, in mm?

Practice

For a tube with A_k = 117,216 mm² carrying T_Ed = 45 kNm, what is the shear flow τt, in N/mm?

Practice

A section has T_Ed/T_Rd,max = 0.45 and V_Ed/V_Rd,max = 0.40. Is the interaction satisfied? Give the utilisation.

Check yourself

Why must links for torsion be closed, when links for shear need not be?

Summary

Ask first whether the torsion is equilibrium or compatibility
Equilibrium torsion must be designed for in full; compatibility torsion may be ignored for strength
Torsional stiffness collapses on cracking, which is WHY compatibility torsion relieves itself
A solid section is analysed as an equivalent thin-walled tube — the core does little
Bredt: τt = T_Ed/(2A_k), with A_k to the wall CENTRELINE, and it is pure equilibrium
Links must be CLOSED, because the shear flow is continuous round the tube
Longitudinal steel is required too, and it is large — not a detailing afterthought
Torsion and shear crush the same struts, so their utilisations ADD

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint