Checks a rectangular or circular column with its bars in shear to SIA 262:2025: ties, compression field and tension chord with a variable-angle stress field and the column load, and the detailing of the ties. Every step shows its formula and the clause it comes from.
What it checks
- Lever arm: z = 0.9 · d, with d to the centroid of the tension bars, or entered (4.3.3.4.2).
- Check 1, ties: Vd ≤ VRd,s = Asw/s · z · fyd · cotα, 4.3.3.4.3 (45). Circular hoops and spirals count for 70 % of rectangular ties of the same depth (4.3.3.4.4).
- Check 2, compression field: Vd ≤ VRd,c = bw · z · kc · fcd · sinα · cosα, 4.3.3.4.6 (47).
- Check 3, tension chord: FtVd = Vd · (cotα − cotβ), 4.3.3.4.11 (51), half to each chord. The share of the column load is Nd,t = Nd · (xc − (d − z))/z, and the chord is checked as Md/z + Nd,t + FtVd/2 ≤ As,t · fyd (4.3.3.4.10).
- Minimum shear reinforcement: ρw = Asw/(s · bw) ≥ ρw,min = 0.001 · √(fck/30) · 500/fyk, 4.3.3.1.1 and 5.5.2.2 (116).
- Tie detailing: s ≤ 15 · Ømin, s ≤ amin and s ≤ 300 mm, 5.5.4.7 (117); Øsw ≥ Øsl,max/3 (5.5.4.9).
What you enter and what you get
You choose a rectangle or a circle, the direction of the shear (along y or along x in the drawing) and the face put in tension by the moment. You click the bars into the drawing, place a whole row at once, or edit them in a table. You enter the materials, the ties (rectangular ties or links, or circular hoops or spirals), the resistance factors, Vd, Md and the column load Nd, the level of approximation and the condition of the column. The angle α is searched or entered.
You get the chords, d and z, each resistance and check with its utilisation and clause, and the distance z · cotα from the support edge at which Vd may be taken (4.3.3.4.1). With no ties chosen, the module shows the reinforcement needed.
Assumptions and limits
- Nd is positive in compression and acts at the centroid of the section. A tension gives αmin = 40°.
- The web is the width across the shear. The bars on each side of the centroid form the two chords; bars on the centroid line belong to neither and are not counted.
- αmin ≤ α ≤ 45°, 4.3.3.3.2 (42). Level 1 fixes αmin (30°, 25° with significant axial compression, 40° with plastic chord deformations) and kc (0.55, or 0.4 with plastic chord deformations, 4.2.1.7). Level 2 takes kc from the strains of the chords, equations (28) and (49), and allows a flatter field, down to 18°.
- Halve the tie spacing where loads are introduced, where the section changes and over laps of the longitudinal bars (5.5.4.9).
- Struts next to the fans over supports, at large loads and at indirect supports need a separate stress field (4.3.3.3.4, 4.3.3.3.6).
Worked example
A 400 × 600 mm column in fck = 30 N/mm² with eight Ø20 bars and Ø10 two-leg ties at 150 mm, shear along the 600 mm depth, under Vd = 350 kN, Md = 250 kNm and Nd = 1,500 kN in compression, at level 1. These are the module's own results for its worked example, computed when this page was built.
Inputs| Input | Value | Ref. |
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| Shape of the section | Rectangle | |
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| Width b | 400 mm | |
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| Depth h | 600 mm | |
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| Shear force acts | Along y — vertical in the drawing | |
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| Face in tension under the moment | Bottom face | |
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| Steel in the section | {"b":[[50,50,20],[200,50,20],[350,50,20],[50,550,20],[200,550,20],[350,550,20],[50,300,20],[350,300,20]],"p":[],"v":[]} | |
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| Lever arm of the internal forces z | 0.9·d (4.3.3.4.2) | § 4.3.3.4.2 |
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| Characteristic compressive strength fck | 30 N/mm² | § 3.1.2.2.2 |
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| Characteristic yield strength of the longitudinal bars fyk | 500 N/mm² | |
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| Shear reinforcement | Rectangular ties or links | |
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| Bar diameter Øsw | 10 mm | |
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| Legs across the shear plane nsw | 2 | |
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| Spacing along the column s | 150 mm | |
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| Characteristic yield strength of the shear reinforcement fyk,w | 500 N/mm² | |
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| Resistance factor for concrete γc | 1.5 | § 2.4.2.6 |
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| Resistance factor for steel γs | 1.15 | § 2.4.2.6 |
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| Duration factor ηt | 1.0 — both conditions of 4.2.1.3 met | § 4.2.1.3 |
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| Design shear force Vd | 350 kN | § 4.3.3.4.1 |
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| Bending moment acting with Vd Md | 250 kNm | |
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| Axial force acting with Vd (compression positive) Nd | 1500 kN | |
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| Level of approximation | Level 1 — fixed limits | § 4.3.3.3.2 |
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| Condition of the column αmin | Normal case — αmin = 30° | § 4.3.3.3.2 |
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| Inclination of the compression field α | Chosen for the largest resistance | § 4.3.3.3.2 |
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Calculation: Concrete section| Step | Value | Ref. |
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| Gross concrete area | Ac = 240,000 mm² | |
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| Depth of the section in the direction of the shear | h = 600 mm | |
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| Web widththe width across the shear | bw = 400 mm | § 4.3.3.3.1 |
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| Depth of the centroid from the compression facebars on the bottom side of this line form the tension chord | xc = 300 mm | |
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| Longitudinal bars | ns = 8Ø20 | |
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Calculation: Chords| Step | Value | Ref. |
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| Bars in the tension chord | As,t = 942 mm² | |
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| Effective depth of the tension steelΣ A·t / Σ A over the tension chordmeasured from the compression face | d = 550 mm | § 4.3.3.4.2 |
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| Steel in the other chord | As,c = 942 mm² | |
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| Lever arm of the internal forces0.9·d | z = 495 mm | § 4.3.3.4.2 |
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Calculation: Shear reinforcement| Step | Value | Ref. |
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| Area of one setnsw·π·Øsw²/4 | Asw = 157 mm² | |
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| Area per unit length | Asw/s = 1.047 mm²/mm | |
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| Minimum shear reinforcement ratio0.001·√(fck/30)·500/fyk | ρw,min = 0.100 % | § 5.5.2.2 (116) |
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| Minimum area per unit lengthρw,min·bw, bw ≤ 400 mm | Asw,min/s = 0.400 mm²/mm | § 5.5.2.2 |
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Calculation: Check — minimum shear reinforcement| Step | Value | Ref. |
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| Shear reinforcement ratioρw | η = 0.382 · Pass38 % of the capacity used | § 4.3.3.1.1, 5.5.2.2 (116) |
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Calculation: Check — spacing and diameter of the ties| Step | Value | Ref. |
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| Spacing of the tiess | η = 0.500 · Pass50 % of the capacity used | § 5.5.4.7 (117) |
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| Diameter of the tiesØsw | η = 0.667 · Pass67 % of the capacity used | § 5.5.4.9 |
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Calculation: Design strengths| Step | Value | Ref. |
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| Design compressive strength of the concreteηfc·ηt·fck/γc, ηfc = (40/fck)^(1/3) ≤ 1 | fcd = 20.0 N/mm² | § 2.4.2.3 (2) |
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| Design yield strength of the shear reinforcementfyk,w/γs | fyd,w = 434.8 N/mm² | § 2.4.2.5 (4) |
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| Design yield strength of the longitudinal barsfyk/γs | fyd = 434.8 N/mm² | § 2.4.2.5 (4) |
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Calculation: Compression field| Step | Value | Ref. |
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| Smallest angle of the compression fieldnormal case | αmin = 30 ° | § 4.3.3.3.2 |
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| Reduction factor for the concrete strengthweb with reinforcement skew to the compression | kc = 0.55 | § 4.2.1.7 |
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| Angle of the compression fieldlargest resistance between αmin and 45° | α = 30.0 ° | § 4.3.3.3.2 (42) |
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| Distance of the section that governs the stirrups from the support edgenear a direct support or a large load, Vd may be taken at this distance | z·cotα = 857 mm | § 4.3.3.4.1 |
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Calculation: Check 1 — shear reinforcement: V_d ≤ V_Rd,s| Step | Value | Ref. |
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| Resistance of the shear reinforcementAsw/s·z·fyd·cotα | VRd,s = 390.4 kN | § 4.3.3.4.3 (45) |
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| Design shear against the shear reinforcementVd/VRd,s | η = 0.897 · Pass90 % of the capacity used | § 4.3.3.4.3 (45) |
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Calculation: Check 2 — compression field: V_d ≤ V_Rd,c| Step | Value | Ref. |
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| Resistance of the compression fieldbw·z·kc·fcd·sinα·cosα | VRd,c = 943.1 kN | § 4.3.3.4.6 (47) |
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| Design shear against the compression fieldVd/VRd,c | η = 0.371 · Pass37 % of the capacity used | § 4.3.3.4.6 (47) |
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Calculation: Check 3 — tension chord with the shear| Step | Value | Ref. |
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| Chord force from the shearVd·(cotα − cotβ)half to each chord (4.3.3.4.11) | FtVd = 606.2 kN | § 4.3.3.4.11 (51) |
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| Share of the axial forceNd·(xc − (d − z))/zNd at the centroid of the section | Nd,t = -742.4 kN | |
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| Tension chordFt,d/Ft,Rd | η = 0.160 · Pass16 % of the capacity used | § 4.3.3.4.10 |
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| Force in the compression chordMd/z − Nd,c − FtVd/2compression: taken by the concrete and the bars of the chord | Fc,d = 959.5 kN | |
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How this module is verified
Verified clause by clause against SIA 262:2025 and against independent calculations used on real projects, before release.
Questions
How does the column load enter the check?
Through the tension chord. A compression Nd gives a negative share Nd,t, which relieves the chord. In the worked example Nd,t = −742 kN.
Why are two of my bars not counted?
They lie on the centroid line of the section, so they belong to neither chord. The module warns when this happens.
Can I print the calculation?
The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.
Related
Norm: SIA 262:2025 · Page updated: 29 October 2026