Checks a tension member of any shape, with its bars and prestressing units, in shear to SIA 262:2025: stirrups, compression field and both chords with a variable-angle stress field, and the detailing of the stirrups. Every step shows its formula and the clause it comes from.

What it checks

Lever arm: a tie has no compression zone, so z is the distance between the centroids of the two groups of steel, or entered (4.3.3.4.2).
Compression field angle: α_min ≤ α ≤ 45°, 4.3.3.3.2 (42), with α_min = 40° because the tie is in tension, and k_c from 4.2.1.7.
Check 1, stirrups: V_d ≤ V_Rd,s = A_sw/s · z · f_yd · cotα, 4.3.3.4.3 (45).
Check 2, compression field: V_d ≤ V_Rd,c = b_w · z · k_c · f_cd · sinα · cosα, 4.3.3.4.6 (47).
Check 3, tension chord: F_tVd = V_d · (cotα − cotβ), 4.3.3.4.11 (51), half to each chord. The share of the tension is N_d,t = N_d · (x_c − (d − z))/z, and the chord is checked as M_d/z + N_d,t + F_tVd/2 ≤ A_s,t · f_yd (4.3.3.4.10).
Check 4, the other chord in tension: −M_d/z + N_d,c + F_tVd/2 ≤ A_s,c · f_yd (4.3.3.4.10).
Detailing: ρ_w = A_sw/(s · b_w) ≥ ρ_w,min = 0.001 · √(f_ck/30) · 500/f_yk, 5.5.2.2 (116), and s ≤ 25 · Ø_sw (5.5.2.2).

What you enter and what you get

You choose a rectangle, a circle or a polygon drawn by clicking its corners, the direction of the shear (along y or along x in the drawing) and the face put in tension by the moment. You click the bars and prestressing units into the drawing, or edit them in a table. You enter the materials, the stirrups, the resistance factors, V_d, M_d and the tension N_d, the level of approximation and whether plastic chord deformations are expected. The angle α is searched or entered.

You get both chords, d and z, each resistance and check with its utilisation and clause, and the distance z · cotα from the support edge at which V_d may be taken (4.3.3.4.1). With no stirrups chosen, the module shows the reinforcement needed.

Assumptions and limits

N_d is the tension in the tie in the same combination, acting at the centroid of the section.
The web is the width across the shear. The steel on each side of the centroid forms a chord; bars on the centroid line belong to neither and are not counted.
α_min = 40°, or 30° if the prestress keeps the concrete compressed. With plastic chord deformations expected, k_c = 0.4 instead of 0.55 (4.2.1.7).
Level 2 takes k_c from the strains of the chords, equations (28) and (49), and allows a flatter field, down to 18°.
Struts next to the fans over supports, at large loads and at indirect supports need a separate stress field (4.3.3.3.4, 4.3.3.3.6).

Worked example

A 400 × 400 mm tie in f_ck = 30 N/mm² with eight Ø26 bars and Ø10 two-leg stirrups at 125 mm, under V_d = 180 kN, M_d = 60 kNm and a tension N_d = 600 kN, at level 1 with elastic chords. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Shape of the section	Rectangle
Width b	400 mm
Depth h	400 mm
Shear force acts	Along y — vertical in the drawing
Face in tension under the moment	Bottom face
Steel in the section	{"b":[[50,50,26],[200,50,26],[350,50,26],[50,350,26],[200,350,26],[350,350,26],[50,200,26],[350,200,26]],"p":[],"v":[]}
Lever arm of the internal forces z	Distance between the two groups of steel	§ 4.3.3.4.2
Characteristic compressive strength f_ck	30 N/mm²	§ 3.1.2.2.2
Characteristic yield strength of the longitudinal bars f_yk	500 N/mm²
Shear reinforcement	Stirrups
Bar diameter Ø_sw	10 mm
Legs across the shear plane n_sw	2
Spacing along the tie s	125 mm
Direction of the stirrups β	Perpendicular to the axis (β = 90°)	§ 4.3.3.4.3
Characteristic yield strength of the shear reinforcement f_yk,w	500 N/mm²
Resistance factor for concrete γ_c	1.5	§ 2.4.2.6
Resistance factor for steel γ_s	1.15	§ 2.4.2.6
Duration factor η_t	1.0 — both conditions of 4.2.1.3 met	§ 4.2.1.3
Design shear force V_d	180 kN	§ 4.3.3.4.1
Bending moment acting with V_d M_d	60 kNm
Tensile force acting with V_d N_d	600 kN
Level of approximation	Level 1 — fixed limits	§ 4.3.3.3.2
Condition of the web α_min	Chords remain elastic	§ 4.3.3.3.2
Inclination of the compression field α	Chosen for the largest resistance	§ 4.3.3.3.2

Calculation: Concrete section
Step	Value	Ref.
Gross concrete area	A_c = 160,000 mm²
Depth of the section in the direction of the shear	h = 400 mm
Web widththe width across the shear	b_w = 400 mm	§ 4.3.3.3.1
Depth of the centroid from the compression facebars on the bottom side of this line form the tension chord	x_c = 200 mm
Longitudinal bars	n_s = 8Ø26

Calculation: Chords
Step	Value	Ref.
Bars in the tension chord	A_s,t = 1,593 mm²
Effective depth of the tension steelΣ A·t / Σ A over the tension chordmeasured from the compression face	d = 350 mm	§ 4.3.3.4.2
Steel in the other chord	A_s,c = 1,593 mm²
Lever arm of the internal forcesbetween the centroids of the two groups of steel	z = 300 mm	§ 4.3.3.4.2

Calculation: Shear reinforcement
Step	Value	Ref.
Area of one setn_sw·π·Ø_sw²/4	A_sw = 157 mm²
Area per unit length	A_sw/s = 1.257 mm²/mm
Minimum shear reinforcement ratio0.001·√(f_ck/30)·500/f_yk	ρ_w,min = 0.100 %	§ 5.5.2.2 (116)
Minimum area per unit lengthρ_w,min·b_w, b_w ≤ 400 mm	A_sw,min/s = 0.400 mm²/mm	§ 5.5.2.2

Calculation: Check — minimum shear reinforcement
Step	Value	Ref.
Shear reinforcement ratioρ_w	η = 0.318 · Pass32 % of the capacity used	§ 5.5.2.2 (116)

Calculation: Check — spacing of the stirrups
Step	Value	Ref.
Spacing of the stirrupss	η = 0.500 · Pass50 % of the capacity used	§ 5.5.2.2

Calculation: Design strengths
Step	Value	Ref.
Design compressive strength of the concreteη_fc·η_t·f_ck/γ_c, η_fc = (40/f_ck)^(1/3) ≤ 1	f_cd = 20.0 N/mm²	§ 2.4.2.3 (2)
Design yield strength of the shear reinforcementf_yk,w/γ_s	f_yd,w = 434.8 N/mm²	§ 2.4.2.5 (4)
Design yield strength of the longitudinal barsf_yk/γ_s	f_yd = 434.8 N/mm²	§ 2.4.2.5 (4)

Calculation: Compression field
Step	Value	Ref.
Smallest angle of the compression fieldthe tie is in tension	α_min = 40 °	§ 4.3.3.3.2
Reduction factor for the concrete strengthweb with reinforcement skew to the compression	k_c = 0.55	§ 4.2.1.7
Angle of the compression fieldlargest resistance between α_min and 45°	α = 40.0 °	§ 4.3.3.3.2 (42)
Distance of the section that governs the stirrups from the support edgenear a direct support or a large load, V_d may be taken at this distance	z·cotα = 358 mm	§ 4.3.3.4.1

Calculation: Check 1 — shear reinforcement: V_d ≤ V_Rd,s
Step	Value	Ref.
Resistance of the shear reinforcementA_sw/s·z·f_yd·cotα	V_Rd,s = 195.3 kN	§ 4.3.3.4.3 (45)
Design shear against the shear reinforcementV_d/V_Rd,s	η = 0.921 · Pass92 % of the capacity used	§ 4.3.3.4.3 (45)

Calculation: Check 2 — compression field: V_d ≤ V_Rd,c
Step	Value	Ref.
Resistance of the compression fieldb_w·z·k_c·f_cd·sinα·cosα	V_Rd,c = 650.0 kN	§ 4.3.3.4.6 (47)
Design shear against the compression fieldV_d/V_Rd,c	η = 0.277 · Pass28 % of the capacity used	§ 4.3.3.4.6 (47)

Calculation: Check 3 — tension chord with the shear
Step	Value	Ref.
Chord force from the shearV_d·(cotα − cotβ)half to each chord (4.3.3.4.11)	F_tVd = 214.5 kN	§ 4.3.3.4.11 (51)
Share of the axial forceN_d·(x_c − (d − z))/zN_d at the centroid of the section	N_d,t = 300.0 kN
Tension chordF_t,d/F_t,Rd	η = 0.877 · Pass88 % of the capacity used	§ 4.3.3.4.10

Calculation: Check 4 — the other chord, in tension
Step	Value	Ref.
Other chord in tensionF_c,d/F_c,Rd	η = 0.299 · Pass30 % of the capacity used	§ 4.3.3.4.10

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 are both chords checked in tension?

The axial tension loads both groups of steel, and each takes half of F_tVd. Check 3 covers the chord that the moment puts in tension, check 4 the other one, with −M_d/z in place of M_d/z.

Why is the compression field at 40° in the worked example?

At level 1, a member in axial tension has α_min = 40° (4.3.3.3.2), or 30° if the prestress keeps the concrete compressed. The angle giving the largest resistance is then α_min itself.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

Norm: SIA 262:2025 · Page updated: 29 October 2026

Reinforced or prestressed concrete tie in shear to SIA 262

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