Concrete wall spanning between two supports, strut-and-tie model to SIA 262
Checks a reinforced concrete wall that spans between two supports, such as columns or walls below, and is loaded on its top edge, to SIA 262:2025, with a strut-and-tie model: the top chord, the tie in the bottom of the wall, the struts and the nodes. 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 wall that spans between two supports, such as columns or walls below, and is loaded on its top edge, to SIA 262:2025, with a strut-and-tie model: the top chord, the tie in the bottom of the wall, the struts and the nodes. Every step shows its formula and the clause it comes from.
What it checks
- The wall as a deep beam: h/l, the bottom tie As,T = nT · π · ØT²/4, d = h − yT, and the ratios ρh and ρv of the bars on both faces, for example ρh = 2 · π · Øh²/4/(sh · bw).
- Design strengths: fcd, 2.4.2.3 (2), with ηfc = (40/fck)^(1/3) ≤ 1, 4.2.1.2 (26); fyd, 2.4.2.5 (4); the strut strength kc · fcd and fcd alone for the nodes and the chord (4.2.1.7).
- Forces: Rd at each support and Md = Rd · a at the load.
- Check 1, top chord: Rd · a ≤ bw · fcd · d²/2 (4.2.1.7). The chord height xc, the lever arm z = d − xc/2, the strut angle θ = atan(z/a), Dd = Rd/sinθ and Td = Rd · a/z follow.
- Check 2, bottom tie: Td ≤ As,T · fyd (2.4.2.5).
- Check 3, support node: the bearing stress Rd/(as · bw) ≤ fcd, and the strut stress Dd/(ws · bw) ≤ kc · fcd, with ws = as · sinθ + u · cosθ and u = 2 · yT (4.2.1.7).
- Check 4, load node: Fd/(aF · bw) ≤ fcd and Dd/(wF · bw) ≤ kc · fcd, with wF = aF · sinθ + xc · cosθ (4.2.1.7).
What you enter and what you get
For the wall you enter the span l between the support centres, its height h and thickness bw, and the length as of each bearing (plate, column or wall below) along the span. The loads on the top edge are two equal point loads placed symmetrically, one point load at mid-span, or a uniform load, with a and the plate length aF. You enter the bottom tie (nT, ØT, yT) and the distributed bars of the wall, none or on both faces (Øh, sh, Øv, sv). Then the materials, the resistance factors and ηt, Fd, and the state of the struts.
You get the forces in the chord, the tie and the struts, the widths of the struts at both nodes, and each check with its utilisation and clause. With the Basic plan, it prints as an A4 calculation note.
Assumptions and limits
- The wall carries its loads on the top edge between two supports.
- Fd is the ULS design value of each load. Add the self-weight of the wall if it matters.
- The support node is u = 2 · yT high, yT measured above the bottom face.
- The diagonal struts spread between the nodes and are crossed by cracks along them: kc = 0.8 with bars across them, 0.55 with bars skew to them, 1.0 uncracked in pure compression.
- The tie is constant over the span. Anchor it behind the inner face of each support for Td (anchorage module, 5.2.5).
- The amount of the distributed bars follows from the crack requirements (4.4.2).
Worked example
A wall 3,000 mm high and 300 mm thick spanning 6,000 mm between two 400 mm supports, with two symmetric loads Fd = 1,200 kN at a = 2,000 mm on 300 mm plates, eight Ø20 bars in the tie 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
Can the load hang from the bottom of the wall?
Only once it is lifted: the module takes the loads on the top edge, and a load hung from the bottom edge must first be lifted to the top by vertical bars.
What do the bars on both faces of the wall do?
They hold the cracks along the struts together and spread the tie force, which lets the struts carry kc = 0.8 (4.2.1.7). Run the horizontal bars into the supports.
Which plan includes this module?
The Basic plan, with its calculation note.
Related
- Concrete deep beam, strut-and-tie model
- Concrete shear wall
- Concrete walls: all modules
- SIA 262: modules and clauses
Norm: SIA 262:2025 · Page updated: 5 November 2026