Shear at the web–flange junction of a concrete T-beam to SIA 262
Checks the longitudinal shear between the web and each flange overhang of a reinforced concrete T- or L-beam, with the flange compressed or in tension, to SIA 262:2025: the struts and the transverse bars of the stress field in the flange, and the minimum transverse reinforcement. 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.
About this check
clauses, worked example and questions
Checks the longitudinal shear between the web and each flange overhang of a reinforced concrete T- or L-beam, with the flange compressed or in tension, to SIA 262:2025: the struts and the transverse bars of the stress field in the flange, and the minimum transverse reinforcement. Every step shows its formula and the clause it comes from.
What it checks
- The overhang: b1 = (beff − bw)/2, with beff from 4.1.3.3.2, equations (19) and (20). The share of the flange force in one overhang is A1/A0 = b1/beff (4.3.4.1.1).
- Design strengths: fcd = ηfc · ηt · fck/γc, 2.4.2.3 (2), with ηfc = (40/fck)^(1/3) ≤ 1, equation (26); fsd = fyk/γs, 2.4.2.5 (4).
- Shear flow through the junction: the chord force changes by ΔFd/Δx = Vd/z; vd = (A1/A0) · ΔFd/Δx (4.3.4.1.1), and the mean stress τd = vd/hf.
- Check 1, struts in the flange: σc = vd/(hf · sinαf · cosαf) ≤ kc · fcd, clause 4.2.1.7, with kc = 0.55 or 0.40.
- Check 2, transverse bars: the tie force td = vd · tanαf (4.3.4.1.2), and asf = vd · tanαf/fsd ≤ as,t + as,b, half at each face of the flange.
- Check 3, minimum reinforcement: as,t + as,b ≥ 0.002 · hf · 1000, that is 0.2 % of the flange section (5.5.2.7).
What you enter and what you get
For the beam you choose a T-section, with overhangs on both sides, or an L-section, with one overhang. You enter h, bw, beff, the flange thickness hf at the junction and the lever arm z from the bending design. You state whether the flange is compressed (sagging moment) or in tension (hogging moment). For the transverse bars crossing the junction: Øt and st of the top layer, Øb and sb of the bottom layer, and the distance c from each face to its layer. Then fck, fyk, γc, γs and ηt. The longitudinal shear comes from Vd, or from the change of moment ΔMd over a length Δx. An optional transverse bending moment md of the flange at the junction (4.3.4.1.3) acts with the shear. Last, the strut angle αf and kc.
You get the shear flow, the strut stress, the bars needed, their values at 25° and 45°, and each check with its utilisation and clause. With the Basic plan, it prints as an A4 calculation note.
Assumptions and limits
- Each junction between the web and an overhang is checked.
- Compressed flange: the compression zone is taken within the flange. If it is deeper, the web takes more and the result stays on the safe side.
- Flange in tension: the overhang carries the tension of the bars placed in it (5.5.2.6).
- In a compressed flange, αf lies between 25° and 45°, 4.3.4.1.2 (52). A smaller angle needs fewer transverse bars and loads the struts more.
- z is about d − hf/2 with the flange in compression, about 0.9 · d otherwise.
- Averaged, the chord force changes by ΔMd/z over the length Δx, for example between a point of zero moment and the largest moment.
- The anchorage is not checked: anchor the transverse bars in the web and beyond the struts in the flange, and the longitudinal bars of the flange for the forces of the stress field (4.3.4.1.4).
Worked example
A T-beam 900 mm deep with a 350 mm web and a 2,400 mm effective flange 200 mm thick, compressed, z = 760 mm, Ø10 at 150 mm in both layers, fck = 30 N/mm², Vd = 850 kN, no transverse moment, αf = 30° and kc = 0.55. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Shape of the section | T-section — overhangs on both sides | |
| Total depth h | 900 mm | |
| Web width bw | 350 mm | |
| Effective width of the flange beff | 2400 mm | § 4.1.3.3.2 |
| Thickness of the flange at the junction hf | 200 mm | |
| Lever arm of the internal forces z | 760 mm | |
| Flange | Compressed — sagging moment, flange on the compression side | § 4.3.4.1.1 |
| Top layer: diameter Øt | 10 mm | |
| Top layer: spacing st | 150 mm | |
| Bottom layer: diameter Øb | 10 mm | |
| Bottom layer: spacing sb | 150 mm | |
| Distance from each face to the centre of its layer c | 35 mm | |
| Characteristic compressive strength fck | 30 N/mm² | § 3.1.2.2.2 |
| Characteristic yield strength fyk | 500 N/mm² | |
| Resistance factor for concrete γc | 1.5 | § 2.4.2.6 |
| Resistance factor for steel γs | 1.15 | § 2.4.2.6 |
| Duration factor ηt | 1.0 — both conditions of 4.2.1.3 met | § 4.2.1.3 |
| Longitudinal shear | From the design shear force Vd | |
| Design shear force of the beam Vd | 850 kN | |
| Angle of the struts to the beam axis αf | 30 ° | § 4.3.4.1.2 (52) |
| Reduction of the concrete strength kc | 0.55 — reinforcement oblique to the struts | § 4.2.1.7 |
| Step | Value | Ref. |
|---|---|---|
| Width of one overhang(beff − bw)/2 | b1 = 1,025 mm | |
| Share of the flange force in one overhangb1/beffcompression zone within the flange; deeper, the web takes more and this stays on the safe side | A1/A0 = 0.427 | § 4.3.4.1.1 |
| Step | Value | Ref. |
|---|---|---|
| Top layerπ·Øt²/4·1000/st | as,t = 524 mm²/m | |
| Bottom layerπ·Øb²/4·1000/sb | as,b = 524 mm²/m | |
| Both layers | as,f = 1,047 mm²/m |
| Step | Value | Ref. |
|---|---|---|
| Change of the chord force per metreVd/z | ΔFd/Δx = 1,118.4 kN/m | |
| Shear flow through the junction(A1/A0)·ΔFd/Δx | vd = 477.7 kN/m | § 4.3.4.1.1 |
| Mean shear stress in the junctionvd/hf | τd = 2.39 N/mm² | |
| Design strength of the concreteηfc·ηt·fck/γcηfc = (40/fck)^(1/3) ≤ 1, eq. (26) | fcd = 20.00 N/mm² | § 2.4.2.3 (2) |
| Design yield strength of the barsfyk/γs | fsd = 434.8 N/mm² | § 2.4.2.5 (4) |
| Step | Value | Ref. |
|---|---|---|
| Compression in the strutsvd/(hf·sin αf·cos αf)αf = 30.0° | σc = 5.52 N/mm² | |
| Strut compression against the reduced strengthσc/(kc·fcd) | η = 0.501 · Pass50 % of the capacity used | § 4.2.1.7 |
| Step | Value | Ref. |
|---|---|---|
| Tie force per metrevd·tan αf | td = 275.8 kN/m | § 4.3.4.1.2 |
| Transverse bars needed for the shearvd·tan αf/fsdhalf at each face of the flange | asf = 634 mm²/m | |
| Transverse bars against the shearasf/(as,t + as,b) | η = 0.606 · Pass61 % of the capacity used | § 4.3.4.1.2 |
| Step | Value | Ref. |
|---|---|---|
| Transverse bars at least 0.2 % of the flange sectionas,min/as,f | η = 0.382 · Pass38 % of the capacity used | § 5.5.2.7 |
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 does only part of the shear pass through one junction?
With the flange compressed, each overhang carries its share b1/beff of the flange force. In the worked example this is 0.427, so vd = 478 kN/m of the 1,118 kN/m change of the chord force.
Which strut angle should I choose?
Any angle from 25° to 45° in a compressed flange. The module shows both limits. In the worked example, 25° needs 512 mm²/m of bars with 6.2 N/mm² in the struts; 45° needs 1,099 mm²/m with 4.8 N/mm², against kc · fcd = 11.0 N/mm².
Which plan includes this module?
The Basic plan, with its calculation note.
Related
- Concrete section in simple bending
- Concrete beam in shear with stirrups
- Concrete beams, bending: all modules
- SIA 262: modules and clauses
Norm: SIA 262:2025 · Page updated: 5 November 2026