Strut-and-Tie Method - Deep Beam Design (EN 1992-1-1)
The strut-and-tie method (STM) designs the discontinuity regions (D-regions) of concrete structures - deep beams, corbels, pile caps, anchor zones - where plane sections do not remain plane and ordinary beam theory does not apply. The load path is idealised as a truss of concrete compression STRUTS and reinforcement TENSION TIES meeting at NODES. This page explains the method and the EN 1992-1-1 §6.5 verifications behind this deep-beam calculator: the truss geometry, the strut and node design strengths, the tie reinforcement, and the strut-angle limits.
The strut-and-tie method (STM) is the Eurocode tool for designing the parts of a concrete structure where ordinary beam theory breaks down. This page explains the method and the EN 1992-1-1 §6.5 checks behind the deep-beam calculator: how the truss is built, and how the struts, ties and nodes are verified.
B-regions and D-regions
A reinforced-concrete member can be split into B-regions ("Bernoulli", where plane sections stay plane and normal bending/shear design applies) and D-regions("discontinuity", where a concentrated load, a support, an opening or an abrupt change of section disturbs the strain field). D-regions extend roughly one member depth from the disturbance. In a D-region the strain is non-linear, so STM - a truss idealisation of the force flow - is used instead. Deep beams, corbels, pile caps, nibs and anchor zones are all D-regions.
The deep beam as a truss
A deep beam (span-to-depth ratio l/h small, roughly ≤3for a simply-supported span per EN 1992-1-1 §5.3.1) carries load mainly by direct strut action, not bending. For a single central top load P, the simplest model is a triangular truss: two inclined compression struts from under the load to each support, and a horizontal tension tie along the bottom between the supports.
Deep-beam strut-and-tie model (central load)
By statics (symmetric):
RA=RB=P/2
Fstrut=sinθRA,Ftie=RAcotθ=RAzl/2
where θ is the strut inclination from the horizontal and zis the internal lever arm between the top load node and the bottom tie (commonly taken as about0.6-0.8h). A larger lever arm steepens the strut and reduces the tie force.
Strut design (§6.5.2)
The design stress in a strut must not exceed σRd,max. With no transverse tension σRd,max=fcd; for a strut in cracked concrete or with transverse tension - the usual "bottle" strut in a deep beam - the limit is reduced:
σRd,max=0.6ν′fcd,ν′=1−fck/250
The acting stress is the strut force over its cross-section: σ=Fstrut/(bws), where b is the member thickness and ws is the strut width set by the node geometry (the bearing length and tie depth projected onto the strut axis).
Node design (§6.5.4)
Nodes are classified by what meets there, and each has a different pressure limitσRd,max=kν′fcd:
Node type
σRd,max
Clause
CCC (struts only)
k1ν′fcd,k1=1.0
EN 1992-1-1 §6.5.4 (6.60)
CCT (one tie)
k2ν′fcd,k2=0.85
EN 1992-1-1 §6.5.4 (6.61)
CTT (two+ ties)
k3ν′fcd,k3=0.75
EN 1992-1-1 §6.5.4 (6.62)
CCC and CCT nodes
CCC - bounded only by struts (e.g. under a concentrated load); highest limit, k1 = 1.0.
CCT - one anchored tie (e.g. a support node: strut + reaction + bottom tie); k2 = 0.85.
CTT - two or more anchored ties; k3 = 0.75.
The reducing factor reflects that anchored ties cause transverse tension that weakens the node. For the deep beam, the support node is a CCT node and the bearing pressureσ=RA/(ba) (with bearing length a) is checked against k2ν′fcd.
Tie design (§6.5.3)
A tie is reinforcement, so the required area is simply:
As,req=fydFtie,fyd=fyk/γs
Critically, the tie must be fully anchored for its force at the nodes. In a deep beam the full tie force is present right at the support, so the bottom reinforcement has to be developed past the support node (often with bends, hooks or anchor plates) - not merely lapped near midspan.
Strut-angle limits
The strut inclination is kept within sensible limits so the truss is realistic - by analogy with the variable-strut-inclination shear method (§6.2.3), 1.0≤cotθ≤2.5, i.e. θ between about 21.8∘ and45∘. Too shallow a strut implies an unrealistically large tie force and a flat compression field; too steep over-stresses the concrete. If the geometry pushesθ outside this band, a single direct strut is not a realistic load path and the model or member proportions should be reconsidered.
Why the method is safe
STM is a lower-bound (static) plasticity method: if a valid equilibrium truss can be found in which every strut, tie and node is adequate, the real capacity is at least that of the model. The result does depend on the truss chosen - good practice is to follow the elastic stress trajectories reasonably closely so the structure does not need excessive deformation to mobilise the assumed load path, and to add minimum crack-control reinforcement alongside the ties.
Frequently asked questions
The strut-and-tie method (STM) is a lower-bound (safe) plasticity method for designing the discontinuity regions (D-regions) of reinforced concrete - deep beams, corbels, pile caps, nibs, openings and anchor zones - where the strain distribution is non-linear and ordinary beam (Bernoulli) theory does not apply. The flow of forces is idealised as a pin-jointed truss: concrete compression members (struts), reinforcement tension members (ties), and the regions where they meet (nodes). If a valid equilibrium truss can be found and every strut, tie and node is adequate, the design is safe by the lower-bound theorem. EN 1992-1-1 §6.5 gives the rules.
Use STM for D-regions: places where a static or geometric discontinuity disturbs the linear strain distribution - within about one member depth of concentrated loads, supports, openings or abrupt changes of section. The classic case is a deep beam: when the span-to-depth ratio l/h is small (roughly l/h ≤ 3 for a simply-supported span per EN 1992-1-1 §5.3.1), the beam carries load mainly by direct strut action to the supports rather than by bending, so a strut-and-tie model is appropriate. For slender beams (B-regions) ordinary bending/shear design applies.
For a simply-supported deep beam with a single central top load P, the simplest model is a triangular truss: two inclined compression struts run from under the load down to each support, and a horizontal tension tie connects the two supports along the bottom. By statics the reactions are P/2 each; the strut force is (P/2)/sinθ and the tie force is (P/2)·cotθ, where θ is the strut inclination from the horizontal and depends on the internal lever arm z between the top load node and the bottom tie (often taken as about 0.6-0.8 of the depth). The bottom tie is the main flexural reinforcement, which must be fully anchored past the supports.
The design compressive stress in a strut must not exceed σRd,max. For a strut with no transverse tension, σRd,max = fcd. For a strut in a cracked region or with transverse tension (a "bottle-shaped" strut, which is the usual case in a deep beam), σRd,max = 0.6·ν′·fcd, where ν′ = 1 - fck/250 is the strength reduction factor for cracked concrete. The acting stress is the strut force divided by the strut cross-section (its width at the node times the member thickness). The strut width follows from the node geometry (the bearing length and the tie depth).
Nodes are classified by what meets there. A CCC node is bounded only by struts (compression-compression-compression) - e.g. under a concentrated load with no anchored tie. A CCT node anchors one tie (compression-compression-tension) - e.g. a support node where a strut, the reaction and the bottom tie meet. A CTT node anchors two or more ties. EN 1992-1-1 §6.5.4 limits the node pressure to σRd,max = k·ν′·fcd, with k = k1 = 1.0 (CCC), k2 = 0.85 (CCT) and k3 = 0.75 (CTT) as the recommended values (a National Annex may differ). The reducing k reflects that anchored ties cause transverse tension that weakens the node.
A tie is a reinforcement member, so its required area is simply As,req = F_tie / fyd, where F_tie is the tie force from the truss and fyd = fyk/γs is the design yield strength (γs = 1.15). Crucially, the tie must be anchored for its full force at the nodes - in a deep beam the bottom reinforcement has to be developed (often with bends or plates) past the support node, not just lapped near midspan, because the full tie force is present right at the support.
The strut inclination should sit within sensible limits so the truss model is realistic - by analogy with the variable-strut-inclination shear method (EN 1992-1-1 §6.2.3), cot θ is kept between 1.0 and 2.5, i.e. θ between about 21.8° and 45°. A very shallow strut (small θ, large cot θ) implies a large tie force and an unrealistic flat compression field; a very steep strut over-stresses the concrete. If your geometry pushes θ outside this band, the load is not really being carried by a single direct strut and the model (or the member proportions) should be reconsidered.
STM is a lower-bound (static) plasticity method, so a design based on any valid equilibrium truss that satisfies the strut, tie and node strengths is safe - the real capacity is at least what the model gives. However, the result depends on the truss you choose: different but equally valid models give different reinforcement. Good practice is to choose a truss that follows the elastic stress trajectories reasonably closely (so the structure does not need excessive deformation to reach the assumed load path) and to provide minimum crack-control reinforcement in addition to the ties.
Ready to design a deep beam? Build the strut-and-tie truss and run the full EN 1992-1-1 §6.5 checks - struts, ties and nodes - with step-by-step derivations.
