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Reinforced concrete column in torsion to SIA 262

Checks a rectangular or circular reinforced concrete column under torsion, with the shear, the bending and the axial force acting with it, to SIA 262:2025: the fictitious hollow section, the ties, the compression field and the chord forces. Every step shows its formula and the clause it comes from.

Basic SIA 262:2025

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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 or circular reinforced concrete column under torsion, with the shear, the bending and the axial force acting with it, to SIA 262:2025: the fictitious hollow section, the ties, the compression field and the chord forces. 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 enclosed by the polygon through the bars, its perimeter uk, 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.
  • Lever arm: z = 0.9 · d, clause 4.3.3.4.2, or entered.
  • Minimum shear reinforcement: ρw ≥ ρw,min, clauses 4.3.3.1.1 and 5.5.2.2, equation (116), with bw taken as at most 400 mm.
  • Ties: spacing s ≤ 15 · Ømin, ≤ amin and ≤ 300 mm, clause 5.5.4.7, equation (117); diameter Øsw ≥ Øsl,max/3, clause 5.5.4.9.
  • Compression field: αmin ≤ α ≤ 45°, clause 4.3.3.3.2, equation (42), with kc from clause 4.2.1.7.
  • Check 1, ties: Vd/VRd,s + Td/TRd,s ≤ 1, clause 4.3.5.5, with VRd,s from 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 from 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).
  • Chord forces: Ft,d = Md/z + Nd,t + FtVd/2 + FtTd/2 and the force Fc,d in the compression chord, with the share of the axial force Nd,t = Nd · (xc − (d − z))/z.

What you enter and what you get

You choose a rectangle or a circle, the direction of Vd in the drawing, and the face in tension under the moment. You click each bar into the drawing, place a row at once with "Bar layer", or edit the table. You enter the shear reinforcement (rectangular ties or links, or circular hoops or spirals) with Øsw, nsw and s, the materials, γc, γs and ηt, then Td, Vd, Md and Nd. You choose level 1 or level 2, the condition of the column, and whether α is searched or entered.

You get the hollow section, the shear in each wall, the resistances, the chord forces, 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

  • Nd is positive in compression. A negative value is a tension, and then αmin = 40°.
  • Only the two outer legs of a set carry the torsion.
  • A circular hoop counts fully for torsion and for 70 % in shear (4.3.3.4.4).
  • Level 2 takes kc from the strains, equations (28) and (49), and allows a field down to 18°.
  • Bars on the centroid line belong to neither chord and are not counted.
  • Halve the tie spacing where loads are introduced, where the section changes and over the laps (5.5.4.9).

Worked example

A 500 × 500 mm column in fck = 30 N/mm² with eight Ø20 bars at the corners and mid-sides, rectangular Ø10 ties with two legs at 150 mm, under Td = 50 kNm, Vd = 150 kN, Md = 120 kNm and Nd = 2,000 kN in compression, 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 b500 mm
Depth h500 mm
Shear force actsAlong y — vertical in the drawing
Face in tension under the momentBottom face
Steel in the section{"b":[[50,50,20],[250,50,20],[450,50,20],[50,450,20],[250,450,20],[450,450,20],[50,250,20],[450,250,20]],"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 reinforcementRectangular ties or links
Bar diameter Øsw10 mm
Legs across the shear plane nsw2
Spacing along the column s150 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 Td50 kNm§ 4.3.5.2
Design shear force acting with Td Vd150 kN§ 4.3.3.4.1
Bending moment acting with Vd Md120 kNm
Axial force acting with Vd (compression positive) Nd2000 kN
Level of approximationLevel 1 — fixed limits§ 4.3.3.3.2
Condition of the column 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 = 250,000 mm²
Depth of the section in the direction of the shearh = 500 mm
Web widththe width across the shearbw = 500 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 = 250 mm
Longitudinal barsns = 8Ø20
Calculation: Chords
StepValueRef.
Bars in the tension chordAs,t = 942 mm²
Effective depth of the tension steelΣ A·t / Σ A over the tension chordmeasured from the compression faced = 450 mm§ 4.3.3.4.2
Steel in the other chordAs,c = 942 mm²
Lever arm of the internal forces0.9·dz = 405 mm§ 4.3.3.4.2
Calculation: Fictitious hollow section
StepValueRef.
Area enclosed by the longitudinal barspolygon through the bars4 cornersAk = 160,000 mm²§ 4.3.5.1
Perimeter of AkΣ ziuk = 1,600 mm
Largest circle inside Akdk = 400 mm§ 4.3.5.3
Thickness of the wallsdk/8tk = 50.0 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 = 1.047 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 mmbw taken as 400 mm (thick web)Asw,min/s = 0.400 mm²/mm§ 5.5.2.2
Calculation: Check — minimum shear reinforcement
StepValueRef.
Shear reinforcement ratioρwη = 0.382 · Pass38 % of the capacity used§ 4.3.3.1.1, 5.5.2.2 (116)
Calculation: Check — spacing and diameter of the ties
StepValueRef.
Spacing of the tiessη = 0.500 · Pass50 % of the capacity used§ 5.5.4.7 (117)
Diameter of the tiesØswη = 0.667 · Pass67 % of the capacity used§ 5.5.4.9
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) = 156.3 N/mm§ 4.3.5.2 (55)
Shear in the wall 1 from the torsionTd/(2·Ak)·zi, zi = 400 mmVd,i = 62.5 kN§ 4.3.5.2 (55)
Shear in the wall 2 from the torsionTd/(2·Ak)·zi, zi = 400 mmVd,i = 62.5 kN§ 4.3.5.2 (55)
Shear in the wall 3 from the torsionTd/(2·Ak)·zi, zi = 400 mmVd,i = 62.5 kN§ 4.3.5.2 (55)
Shear in the wall 4 from the torsionTd/(2·Ak)·zi, zi = 400 mmVd,i = 62.5 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 = 30.0 °§ 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 = 319.4 kN§ 4.3.3.4.3 (45)
Torsional resistance of the stirrups2·Ak·Asw1/s·fyd·cotαthe outer legs carry the shear flowTRd,s = 126.2 kNm§ 4.3.5.3, 4.3.3.4.3 (45)
Stirrups, shear and torsion togetherVd/VRd,s + Td/TRd,sη = 0.866 · Pass87 % 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 = 964.5 kN§ 4.3.3.4.6 (47)
Torsional resistance of the compression field2·Ak·tk·kc·fcd·sinα·cosαTRd,c = 76.2 kNm§ 4.3.5.3, 4.3.3.4.6 (47)
Concrete, shear and torsion togetherVd/VRd,c + Td/TRd,cη = 0.812 · Pass81 % 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 = 433.0 kN§ 4.3.5.4, 4.3.3.4.11 (51)
Per unit length of the perimeterTd·cotα/(2·Ak)ftTd = 270.6 kN/m
Calculation: Tension chord
StepValueRef.
Chord force from the shearVd·cotαhalf to each chordFtVd = 259.8 kN§ 4.3.3.4.11 (51)
Share of the axial forceNd·(xc − (d − z))/zNd at the centroid of the sectionNd,t = -1,012.3 kN
Force in the tension chordMd/z + Nd,t + FtVd/2 + FtTd/2negative: the axial compression keeps this chord compressedFt,d = -369.6 kN
Force in the compression chordMd/z − Nd,c − FtVd/2 − FtTd/2compression: taken by the concrete and the bars of the chordFc,d = 937.5 kN

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 can the force in the tension chord be negative?

The axial compression can outweigh the tension from the moment, the shear and the torsion. A negative Ft,d means that the chord stays compressed.

Can the ties be omitted?

Choose "None" to see the reinforcement needed. Torsion needs closed ties or hoops that follow the perimeter and are anchored round the whole hollow section (4.3.5.4, 5.5.2.4).

Which plan includes this module?

The Basic plan, with its calculation note.

Related

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