Torsion in concrete

What this calculator checks

The tool determines the torsional resistance of a rectangular reinforced concrete section and the reinforcement it requires, to EN 1992-1-1 §6.3. It covers solid and hollow sections, with the torsion–shear interaction.

The model: an equivalent thin-walled tube

The code does not treat the section as a solid body. It replaces it with a closed thin-walled tube, and torsion circulates around its perimeter as a constant shear flow:

\(\tau_{t,i} \cdot t_{ef,i} = \frac{T_{Ed}}{2 A_k}\)

From this follows a fact that surprises on first encounter: the core of the section carries no torsion. A solid beam and a hollow one with the same wall thickness resist practically the same. The concrete in the middle contributes to bending and shear, but for torsion it is ballast.

The tube geometry is built as:

\(t_{ef} = \frac{A}{u}, \qquad A_k = (b - t_{ef})(h - t_{ef}), \qquad u_k = 2\left[(b - t_{ef}) + (h - t_{ef})\right]\)

where \(A\) is the total section area and \(u\) its outer perimeter. \(A_k\) is the area enclosed by the wall centre-lines — hence subtracting one \(t_{ef}\), not two: the centre-line sits \(t_{ef}/2\) from each face.

The effective thickness has two limits. It cannot fall below twice the distance from the face to the axis of the longitudinal reinforcement (clause 6.3.2(1)) — the bars must fit inside the wall. And it cannot exceed the actual wall thickness of a hollow section.

The check that most often ends the calculation

Before sizing reinforcement, expression 6.31 is worth checking:

\(\frac{T_{Ed}}{T_{Rd,c}} + \frac{V_{Ed}}{V_{Rd,c}} \le 1\)

where \(T_{Rd,c} = 2 A_k t_{ef} f_{ctd}\) is the torque at which the section starts to crack in torsion.

If it is satisfied, clause 6.3.2(5) states that no calculated torsion reinforcement is required — the minimum detailing of clause 9.2.1.1 suffices. For ordinary floor beams, where torsion arises from compatibility rather than equilibrium, the condition is often met and the calculation stops here.

Note: the exemption applies only to approximately rectangular solid sections.

Crushing of the struts

\(\frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} \le 1\)

\(T_{Rd,max} = 2 \nu \alpha_{cw} f_{cd} A_k t_{ef} \sin\theta \cos\theta\)

The interaction is linear, not quadratic — torsion and shear draw on the same budget of concrete compression. This is the limit that reinforcement cannot fix: if 6.29 is not satisfied, the only remedies are a larger section or a higher concrete class.

The product \(\sin\theta \cos\theta\) peaks at \(\theta = 45°\), that is \(\cot\theta = 1\). In other words, a 45° strut gives the highest crushing resistance — the opposite of shear, where a large \(\cot\theta\) is favourable.

Reinforcement: the two directions move in opposite ways

\(\sum A_{sl} = \frac{T_{Ed} \cot\theta \; u_k}{2 A_k f_{yd}} \qquad \left(\frac{A_{sw}}{s}\right)_T = \frac{T_{Ed}}{2 A_k f_{ywd} \cot\theta}\)

\(\cot\theta\) appears in the numerator for longitudinal steel and in the denominator for stirrups. A flatter strut (large cot θ) needs fewer stirrups but more longitudinal bars; a 45° strut does the reverse. Choosing θ is an economic decision, and the page exposes it with a slider precisely so the trade-off is visible.

The trap when combining with shear

This is where many go wrong. In torsion, \(A_{sw}\) is the area of a single leg of a closed stirrup — the flow travels around the section, so each wall needs one leg. In shear, \(A_{sw}\) is the area of all vertical legs, usually two.

Combined on a two-leg stirrup:

\(\left(\frac{A_{sw}}{s}\right)_{\text{per leg}} = \left(\frac{A_{sw}}{s}\right)_T + \frac{1}{2}\left(\frac{A_{sw}}{s}\right)_V\)

Adding the two values directly overestimates the requirement by up to 50%. The page shows the three rows separately, precisely to make the operation visible.

One more thing: torsion requires closed stirrups, anchored so the flow can circulate. Open stirrups carry no torsion, however large their area.

Input data

• Actions — \(T_{Ed}\), \(V_{Ed}\) and optionally \(N_{Ed}\) (which enters only \(V_{Rd,c}\)).
• Section — \(b\), \(h\), \(d\), the distance to the bar axis, solid or hollow.
• Material — \(f_{ck}\), \(f_{yk}\), plus \(A_{sl}\) for \(\rho_l\) in \(V_{Rd,c}\).
• cot θ — between 1.0 and 2.5.

Assumptions and limitations

• Rectangular sections only, solid or hollow. T, L and I sections are decomposed into rectangles, with the torque shared in proportion to each one's stiffness — that calculation is not done here.
• Warping torsion in open thin-walled profiles is not covered. There the warping component dominates and the tube model does not apply.
• Interaction with bending is not included in the same expression. The longitudinal torsion reinforcement adds to the bending reinforcement, distributed around the perimeter.
• Detailing rules are not checked: maximum stirrup spacing, minimum area, placement of longitudinal bars at the corners.
• Above C50/60 the expression for \(f_{ctm}\) changes; the page uses the normal-strength form and flags the case.

Related calculators

• Reinforced concrete section — bending and shear, EN 1992
• Punching shear — the other special shear check
• Minimum reinforcement ratios — the detailing minima that remain mandatory