Reinforced concrete deep beam, strut-and-tie model to SIA 262
Checks a reinforced concrete deep beam spanning between two supports and loaded on its top edge, to SIA 262:2025, with a strut-and-tie model: the compression chord, the bottom tie, the struts and the nodes over the supports and under the loads. 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 a reinforced concrete deep beam spanning between two supports and loaded on its top edge, to SIA 262:2025, with a strut-and-tie model: the compression chord, the bottom tie, the struts and the nodes over the supports and under the loads. Every step shows its formula and the clause it comes from.
What it checks
- Geometry: h/l, the tie area As,T = nT · π · ØT²/4 and d = h − yT. Web bars: ρh = 2 · π · Øh²/4/(sh · bw) and ρv likewise.
- Design strengths: ηfc, 4.2.1.2 (26); fcd = ηfc · ηt · fck/γc, 2.4.2.3 (2); fyd = fyk/γs, 2.4.2.5 (4). The struts take kc · fcd, the nodes and the compression chord fcd (clause 4.2.1.7).
- Model: the support reaction Rd and the moment at the load Md = Rd · a, with a from the support centre to the load resultant.
- Check 1, compression chord: Rd · a ≤ bw · fcd · d²/2 (4.2.1.7). Then xc = T/(bw · fcd), z = d − xc/2, θ = atan(z/a), the strut force Dd = Rd/sinθ and the chord and tie force Td = Rd · a/z.
- Check 2, tie: Td ≤ As,T · fyd (2.4.2.5).
- Check 3, node over the support: node height u = 2 · yT, strut width ws = as · sinθ + u · cosθ; Rd/(as · bw) ≤ fcd and Dd/(ws · bw) ≤ kc · fcd (4.2.1.7).
- Check 4, node under the load: wF = aF · sinθ + xc · cosθ; Fd/(aF · bw) ≤ fcd and Dd/(wF · bw) ≤ kc · fcd (4.2.1.7).
What you enter and what you get
For the deep beam you enter the span l between the support centres, the height h, the thickness bw and the length as of each support. You choose the loading: two equal symmetric point loads, one point load at mid-span, or a uniform load on the top edge, with a and the load plate aF. For the tie: nT, ØT and the height yT of its centroid. For the web: no distributed bars, or bars on both faces with Øh, sh, Øv and sv. Then fck, fyk, γc, γs, ηt, the design value Fd of each load and the state of the struts.
You get the forces of the model, the geometry of the nodes and each check with its utilisation and clause. With the Basic plan, it prints as an A4 calculation note.
Assumptions and limits
- The loads act on the top edge. A load hung from the bottom edge must first be lifted to the top by vertical bars.
- Fd comes from the ULS combination; add the self-weight of the deep beam if it matters.
- The node over the support is taken 2 · yT high.
- The tie force is constant from support to support. Its anchorage behind the inner face of each support, with the bars run to the end of the beam, is checked with the anchorage module (5.2.5).
- The struts take kc = 0.8 when cracked along their length with web bars across them, 0.55 with web bars skew to them, and 1.0 when uncracked in pure compression.
- The module gives the ratios of the web bars. Their amount follows from the crack requirements (4.4.2); run the horizontal bars into the supports.
Worked example
A deep beam with l = 6,000 mm, h = 3,000 mm and bw = 300 mm on 400 mm supports, carrying two symmetric loads Fd = 1,200 kN at a = 2,000 mm on 300 mm plates, with eight Ø20 tie bars at yT = 100 mm, Ø10 at 200 mm both ways on both faces, fck = 30 N/mm² and kc = 0.8. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Span between the support centres l | 6000 mm | |
| Height h | 3000 mm | |
| Thickness bw | 300 mm | |
| Length of each support as | 400 mm | |
| Loading | Two equal point loads, symmetric | |
| Distance from a support centre to the load a | 2000 mm | |
| Length of the load plate aF | 300 mm | |
| Number of bars in the tie nT | 8 | |
| Diameter of the tie bars ØT | 20 mm | |
| Height of the centroid of the tie yT | 100 mm | |
| Distributed bars in the web | Bars on both faces | |
| Diameter of the horizontal bars Øh | 10 mm | |
| Spacing of the horizontal bars sh | 200 mm | |
| Diameter of the vertical bars Øv | 10 mm | |
| Spacing of the vertical bars sv | 200 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 |
| Design value of each load Fd | 1200 kN | |
| State of the struts kc | Cracks along the strut, web bars across it — kc = 0.8 | § 4.2.1.7 |
| Step | Value | Ref. |
|---|---|---|
| Height to spana deep beam: plane sections do not stay plane; a strut-and-tie model is needed | h/l = 0.50 | |
| Distance from the support to the load resultantentered | a = 2,000 mm | |
| Area of the tienT·π·ØT²/4 | As,T = 2,513 mm² | |
| Effective depthh − yT | d = 2,900 mm |
| Step | Value | Ref. |
|---|---|---|
| Ratio of the horizontal bars2·π·Øh²/4/(sh·bw) | ρh = 0.26 % | |
| Ratio of the vertical bars2·π·Øv²/4/(sv·bw) | ρv = 0.26 % |
| Step | Value | Ref. |
|---|---|---|
| Factor for the brittleness of strong concrete(40/fck)^(1/3) ≤ 1 | ηfc = 1.000 | § 4.2.1.2 (26) |
| Design compressive strength of the concreteηfc·ηt·fck/γc | fcd = 20.0 N/mm² | § 2.4.2.3 (2) |
| Design yield strength of the barsfyk/γs | fyd = 434.8 N/mm² | § 2.4.2.5 (4) |
| Strength of the strutskc = 0.80the nodes and the compression chord take fcd (kc = 1.0) | kc·fcd = 16.0 N/mm² | § 4.2.1.7 |
| Step | Value | Ref. |
|---|---|---|
| Support reactionFd | Rd = 1,200 kN | |
| Moment at the loadRd·a | Md = 2,400 kNm |
| Step | Value | Ref. |
|---|---|---|
| Moment against the largest the chord allowsMd/Mc,max | η = 0.095 · Pass10 % of the capacity used | § 4.2.1.7 |
| Height of the compression chordT/(bw·fcd) | xc = 141 mm | |
| Lever armd − xc/2, from Rd·a = T·z | z = 2,829 mm | |
| Inclination of the strutsatan(z/a) | θ = 54.7 ° | |
| Force in each strutRd/sinθ | Dd = 1,470 kN | |
| Force in the chord and in the tieRd·a/z | Td = 848 kN |
| Step | Value | Ref. |
|---|---|---|
| Tie force against the bottom barsTd/TRd | η = 0.776 · Pass78 % of the capacity used | § 2.4.2.5 |
| Step | Value | Ref. |
|---|---|---|
| Height of the node2·yT | u = 200 mm | |
| Width of the strut at the nodeas·sinθ + u·cosθ | ws = 442 mm | |
| Bearing stress on the supportσR/fcd | η = 0.500 · Pass50 % of the capacity used | § 4.2.1.7 |
| Stress in the strut at the supportσD/(kc·fcd) | η = 0.693 · Pass69 % of the capacity used | § 4.2.1.7 |
| Step | Value | Ref. |
|---|---|---|
| Width of the strut at the nodeaF·sinθ + xc·cosθ | wF = 327 mm | |
| Bearing stress under the load plateσF/fcd | η = 0.667 · Pass67 % of the capacity used | § 4.2.1.7 |
| Stress in the strut under the loadσD/(kc·fcd) | η = 0.937 · Pass94 % of the capacity used | § 4.2.1.7 |
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 a strut-and-tie model and not a section check?
In a deep beam plane sections do not stay plane, so a strut-and-tie model is needed. The module shows h/l; in the worked example it is 0.5.
What does the chord check limit?
The largest moment the compression chord allows at fcd, bw · fcd · d²/2. Beyond it, no lever arm balances the moment.
Which plan includes this module?
The Basic plan, with its calculation note.
Related
- Concrete wall, strut-and-tie model
- Concrete beam in shear with stirrups
- Concrete beams, deep beams: all modules
- SIA 262: modules and clauses
Norm: SIA 262:2025 · Page updated: 5 November 2026