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Reinforced concrete beam in torsion to SIA 262

Checks a rectangular, box or convex concrete beam under torsion, with the shear and the bending acting with it, to SIA 262:2025: the fictitious hollow section, the stirrups, the compression field and the longitudinal bars. Every step shows its formula and the clause it comes from.

Basic SIA 262:2025

Part of the Basic plan

This module is included in the Basic plan and above, with its calculation note.

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About this check

clauses, worked example and questions

Checks a rectangular, box or convex concrete beam under torsion, with the shear and the bending acting with it, to SIA 262:2025: the fictitious hollow section, the stirrups, the compression field and the longitudinal bars. Every step shows its formula and the clause it comes from.

What it checks

  • Fictitious hollow section, clause 4.3.5.1 and Figure 20: Ak is the area enclosed by the polygon through the longitudinal bars, uk its perimeter, dk the largest circle inside it, and tk = dk/8, clause 4.3.5.3, equation (56).
  • Shear flow, clause 4.3.5.2, equation (55): Vd,i = Td/(2 · Ak) · zi in each wall.
  • Chords: d to the centroid of the tension steel and z = 0.9 · d, clause 4.3.3.4.2, or z entered.
  • Minimum shear reinforcement: ρw ≥ ρw,min = 0.001 · √(fck/30) · 500/fyk, clause 5.5.2.2, equation (116), and stirrup spacing s ≤ 25 · Øsw.
  • Compression field: αmin ≤ α ≤ 45°, clause 4.3.3.3.2, equation (42), with kc from clause 4.2.1.7.
  • Check 1, stirrups: Vd/VRd,s + Td/TRd,s ≤ 1, clause 4.3.5.5, with VRd,s = Asw/s · z · fyd · cotα, equation (45), and TRd,s = 2 · Ak · Asw1/s · fyd · cotα.
  • Check 2, compression field: Vd/VRd,c + Td/TRd,c ≤ 1, clause 4.3.5.5, with VRd,c = bw · z · kc · fcd · sinα · cosα, equation (47), and TRd,c = 2 · Ak · tk · kc · fcd · sinα · cosα.
  • Longitudinal force from the torsion: FtTd = Td · uk · cotα/(2 · Ak), clauses 4.3.5.4 and 4.3.3.4.11, equation (51).
  • Checks 3 and 4, chords: Md/z, the axial force, half of FtVd = Vd · cotα and half of FtTd against As,t · fyd and As,c · fyd, clauses 4.3.3.4.10 and 4.3.5.4.

What you enter and what you get

You choose the shape (rectangle, hollow box, or a convex polygon drawn by clicking its corners) and the face in tension under the moment. You click each reinforcing bar or prestressing unit into the drawing, or edit the table. You enter the closed stirrups Øsw, nsw and s, the materials, γc, γs and ηt, then Td, Vd, Md and Nd. You choose level 1 or level 2 for the compression field, the condition of the web, and whether α is searched or entered.

You get the hollow section, the shear in each wall, the resistances, every check with its utilisation and clause, and a drawing of the section. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

  • A T- or L-beam can be drawn as the polygon of its web.
  • Nd is positive in tension. Prestress is placed in the section, not entered as a force.
  • Only the two outer legs of a set carry the torsion; all nsw legs carry the shear.
  • Level 2 takes kc from the strains of the chords, equations (28) and (49), and allows a field down to 18°.
  • Choose "None" for the stirrups to see the reinforcement needed.
  • Bars on the centroid line belong to neither chord and are not counted.
  • The longitudinal torsion bars go along the walls or at the corners; the stirrups must be closed and anchored (4.3.5.4, 5.5.2.4).

Worked example

A 400 × 700 mm rectangular beam in fck = 30 N/mm², with three Ø26 in the bottom tension chord, two Ø16 at the top and two Ø12 at mid-depth, closed Ø10 stirrups with two legs at 125 mm, under Td = 40 kNm, Vd = 200 kN and a sagging Md = 150 kNm, at level 1 in the normal case. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Shape of the sectionRectangle
Width b400 mm
Depth h700 mm
Face in tensionBottom face in tension (sagging moment)
Steel in the section{"b":[[50,50,26],[200,50,26],[350,50,26],[50,650,16],[350,650,16],[50,350,12],[350,350,12]],"p":[],"v":[]}
Lever arm of the internal forces z0.9·d (4.3.3.4.2)§ 4.3.3.4.2
Thickness of the walls tktk = dk/8 (eq. (56))§ 4.3.5.3 (56)
Characteristic compressive strength fck30 N/mm²§ 3.1.2.2.2
Characteristic yield strength of the longitudinal bars fyk500 N/mm²
Shear reinforcementClosed stirrups
Bar diameter Øsw10 mm
Legs across the shear plane nsw2
Spacing along the beam s125 mm
Characteristic yield strength of the shear reinforcement fyk,w500 N/mm²
Resistance factor for concrete γc1.5§ 2.4.2.6
Resistance factor for steel γs1.15§ 2.4.2.6
Duration factor ηt1.0 — both conditions of 4.2.1.3 met§ 4.2.1.3
Design torsional moment Td40 kNm§ 4.3.5.2
Design shear force acting with Td Vd200 kN§ 4.3.3.4.1
Bending moment acting with Vd Md150 kNm
Level of approximationLevel 1 — fixed limits§ 4.3.3.3.2
Condition of the web αminNormal case — αmin = 30°§ 4.3.3.3.2
Inclination of the compression field αChosen for the largest resistance§ 4.3.3.3.2
Calculation: Concrete section
StepValueRef.
Gross concrete areaAc = 280,000 mm²
Depth of the section in the direction of the shearh = 700 mm
Web widththe width across the shearbw = 400 mm§ 4.3.3.3.1
Depth of the centroid from the compression facebars on the bottom side of this line form the tension chordxc = 350 mm
Longitudinal barsns = 2Ø12 + 2Ø16 + 3Ø26
Calculation: Chords
StepValueRef.
Bars in the tension chordAs,t = 1,593 mm²
Effective depth of the tension steelΣ A·t / Σ A over the tension chordmeasured from the compression faced = 650 mm§ 4.3.3.4.2
Steel in the other chordAs,c = 402 mm²
Lever arm of the internal forces0.9·dz = 585 mm§ 4.3.3.4.2
Calculation: Fictitious hollow section
StepValueRef.
Area enclosed by the longitudinal barspolygon through the bars4 cornersAk = 180,000 mm²§ 4.3.5.1
Perimeter of AkΣ ziuk = 1,800 mm
Largest circle inside Akdk = 300 mm§ 4.3.5.3
Thickness of the wallsdk/8tk = 37.5 mm§ 4.3.5.3 (56)
Calculation: Torsion and shear reinforcement
StepValueRef.
Area of one setnsw·π·Øsw²/4Asw = 157 mm²
Area of one outer leg, for the torsionπ·Øsw²/4Asw1 = 79 mm²
Area per unit lengthAsw/s = 1.257 mm²/mm
Minimum shear reinforcement ratio0.001·√(fck/30)·500/fykρw,min = 0.100 %§ 5.5.2.2 (116)
Minimum area per unit lengthρw,min·bw, bw ≤ 400 mmAsw,min/s = 0.400 mm²/mm§ 5.5.2.2
Calculation: Check — minimum shear reinforcement
StepValueRef.
Shear reinforcement ratioρwη = 0.318 · Pass32 % of the capacity used§ 5.5.2.2 (116)
Calculation: Check — spacing of the stirrups
StepValueRef.
Spacing of the stirrupssη = 0.500 · Pass50 % of the capacity used§ 5.5.2.2
Calculation: Design strengths
StepValueRef.
Design compressive strength of the concreteηfc·ηt·fck/γc, ηfc = (40/fck)^(1/3) ≤ 1fcd = 20.0 N/mm²§ 2.4.2.3 (2)
Design yield strength of the shear reinforcementfyk,w/γsfyd,w = 434.8 N/mm²§ 2.4.2.5 (4)
Design yield strength of the longitudinal barsfyk/γsfyd = 434.8 N/mm²§ 2.4.2.5 (4)
Calculation: Torsion on the fictitious hollow section
StepValueRef.
Shear flow of the torsionuniform round the sectionTd/(2·Ak) = 111.1 N/mm§ 4.3.5.2 (55)
Shear in the wall 1 from the torsionTd/(2·Ak)·zi, zi = 300 mmVd,i = 33.3 kN§ 4.3.5.2 (55)
Shear in the wall 2 from the torsionTd/(2·Ak)·zi, zi = 600 mmVd,i = 66.7 kN§ 4.3.5.2 (55)
Shear in the wall 3 from the torsionTd/(2·Ak)·zi, zi = 300 mmVd,i = 33.3 kN§ 4.3.5.2 (55)
Shear in the wall 4 from the torsionTd/(2·Ak)·zi, zi = 600 mmVd,i = 66.7 kN§ 4.3.5.2 (55)
Calculation: Compression field
StepValueRef.
Smallest angle of the compression fieldnormal caseαmin = 30 °§ 4.3.3.3.2
Reduction factor for the concrete strengthkc = 0.55§ 4.2.1.7
Angle of the compression fieldsmallest utilisation between αmin and 45°α = 35.4 °§ 4.3.3.3.2 (42)
Calculation: Check 1 — stirrups under shear and torsion
StepValueRef.
Shear resistance of the stirrupsAsw/s·z·fyd·cotαVRd,s = 449.8 kN§ 4.3.3.4.3 (45)
Torsional resistance of the stirrups2·Ak·Asw1/s·fyd·cotαthe outer legs carry the shear flowTRd,s = 138.4 kNm§ 4.3.5.3, 4.3.3.4.3 (45)
Stirrups, shear and torsion togetherVd/VRd,s + Td/TRd,sη = 0.734 · Pass73 % of the capacity used§ 4.3.5.5
Calculation: Check 2 — compression field under shear and torsion
StepValueRef.
Shear resistance of the compression fieldbw·z·kc·fcd·sinα·cosαVRd,c = 1,215.4 kN§ 4.3.3.4.6 (47)
Torsional resistance of the compression field2·Ak·tk·kc·fcd·sinα·cosαTRd,c = 70.1 kNm§ 4.3.5.3, 4.3.3.4.6 (47)
Concrete, shear and torsion togetherVd/VRd,c + Td/TRd,cη = 0.735 · Pass74 % of the capacity used§ 4.3.5.5
Calculation: Longitudinal force from the torsion
StepValueRef.
Longitudinal force from the torsionΣ Vd,i·cotα = Td·uk·cotα/(2·Ak)FtTd = 281.4 kN§ 4.3.5.4, 4.3.3.4.11 (51)
Per unit length of the perimeterTd·cotα/(2·Ak)ftTd = 156.3 kN/m
Calculation: Check 3 — tension chord
StepValueRef.
Chord force from the shearVd·cotαhalf to each chordFtVd = 281.4 kN§ 4.3.3.4.11 (51)
Tension chordFt,d/Ft,Rdη = 0.777 · Pass78 % of the capacity used§ 4.3.3.4.10, 4.3.5.4
Calculation: Check 4 — the other chord, in tension
StepValueRef.
Other chord in tensionFc,d/Fc,Rdη = 0.143 · Pass14 % of the capacity used§ 4.3.3.4.10, 4.3.5.4

How this module is verified

Verified clause by clause against SIA 262:2025 and against independent calculations used on real projects, before release.

Questions

Why are shear and torsion added in one check?

The same stirrups and the same compression field take both. Clause 4.3.5.5 sums the two utilisations, for the stirrups and for the concrete separately.

Where does the extra longitudinal steel go?

The force FtTd is spread along the walls or concentrated at the corners (4.3.5.4). The module adds half of it to each chord.

Which plan includes this module?

The Basic plan, with its calculation note.

Related

Norm: SIA 262:2025 · Page updated: 29 October 2026