Strut-and-tie model

What this page calculates

The calculator builds the strut-and-tie model for you in a reinforced concrete D-region — a deep beam, a wall with openings, a corbel, a bearing zone — and works out its capacity. You give the panel, the bearings, the loads and the reinforcement you have detailed; the model is the result, not something you draw beforehand.

The method is lower-bound limit analysis, which is precisely the basis on which EN 1992-1-1 §6.5 permits strut-and-tie design: among all statically admissible stress fields that nowhere exceed the material strengths, it looks for the one carrying the most load.

\(\max \lambda \quad \text{subject to} \quad B_C q_C + B_S q_S = \lambda f\)

The result, the load factor \(\lambda\), is safe by construction: the real capacity is at least that large.

B-regions and D-regions

In a B-region (Bernoulli) plane sections remain plane, so ordinary bending and shear design applies. In a D-region (discontinuity) that assumption fails: near supports, under concentrated loads, at abrupt changes of section, around openings. The practical rule is that a D-region extends a distance equal to the depth of the member, measured from the discontinuity.

There you may no longer design with beam formulas — and that is where the strut-and-tie model comes in.

The two tabs

• CSFM model — the struts (compression) and ties (tension) that actually carry the load, with line thickness proportional to utilisation. Out of tens of thousands of candidate members, fewer than twenty typically remain.
• Stresses — the concrete utilisation map from a second solver, a continuum finite element stress field. It shows how the compression really spreads through the member, not just the idealised path.

The two complement each other: the first gives the capacity, the second gives the picture.

Strut strength

Clause §6.5.2 applies. The default is the value for cracked zones with transverse tension, because a lower-bound model cannot know in advance where that arises:

\(\sigma_{Rd,max} = 0.6 \, \nu' f_{cd}, \qquad \nu' = 1 - \frac{f_{ck}}{250}\)

If you know a strut lies in a zone free of transverse tension, you can switch to the full value \(f_{cd}\).

Input data

• The panel — length, height, thickness; optionally rectangular openings.
• Candidate node spacing — how fine the set is from which the model is chosen.
• The bearings — bearing plates, restrained in x, in y or in both.
• The loads — total forces over a patch, horizontal and vertical.
• The reinforcement — segments with a steel area. Ties exist only where you have placed reinforcement; no steel is invented.
• The materials — concrete class, steel grade, partial factors.

Worked example

A deep beam 4.00 × 2.00 m, 0.30 m thick, C30/37, S500 steel, a 400 kN concentrated load at mid-span on the top edge, and a bottom tie of 10 cm² at 12.5 cm from the soffit.

• \(f_{cd} = 1.00 \cdot 30.0 / 1.50 = 20.00\) MPa
• \(\nu' = 1 - 30/250 = 0.880\)
• \(\sigma_{Rd,max} = 0.6 \cdot 0.880 \cdot 20.00 = 10.56\) MPa
• \(f_{yd} = 500 / 1.15 = 434.8\) MPa

The result is \(\lambda = 2.446\), a capacity of about 978 kN. The bottom tie governs, at a utilisation of 1.000 — the steel yields before the concrete reaches 25% of the strut strength. The resulting model has 19 struts, chosen out of 15,035 candidates.

A modelling trap

Restrain horizontal displacement at a single node. If you restrain x over the whole bearing plate, the beam turns into an arch whose thrust is taken by the support, and the capacity rises artificially because the tie is no longer stressed. It is the most common mistake with this kind of model.

What it does not cover

• Node checks to §6.5.4 are not automated. Stresses on the faces of CCC, CCT and CTT nodes remain the designer's responsibility. For every strut, though, the width it needs is reported, \(w_{req} = |q| / (\sigma_{Rd,max} \, t)\), precisely so that the check can be made; if the required width exceeds the narrowest bearing, you get a warning.
• Anchorage of the ties beyond the node — see anchorage and laps and lap detailing. A tie that cannot be anchored does not exist, however good the model looks.
• Crack widths and the serviceability state — the model is an ultimate limit state check. For SLS see the crack width check.
• Minimum detailing reinforcement — see minimum reinforcement ratios. EN 1992-1-1 §9.7 requires an orthogonal mesh in deep beams whatever the model says.
• The analysis is planar; members behaving three-dimensionally do not belong here.

Assumptions and limitations

• Concrete has no tensile strength. Struts carry compression only.
• The load factor is governed by the reinforcement, not by an assumed strut area. That is deliberate: an area derived from the node spacing would make the result depend on the discretisation, which is not a physical property. Strut strength enters through the required widths, checked separately. You can still cap the struts at a physical width of your choosing.
• The load factor is a lower bound, not an estimate of the real failure load.
• The analysis runs on an external calculation engine.

Related calculations

• Slab analysis by finite elements — the stress field for slabs
• RC section — design in B-regions
• Punching shear — another D-region, handled with closed-form rules
• Crack width check — the serviceability limit state