Checks one-way shear in a reinforced concrete slab strip 1 m wide to SIA 262:2025: the resistance without shear reinforcement from the strain of the bending bars, and a proposed shear reinforcement when it is needed. Every step shows its formula and the clause it comes from.
What it checks
- Effective depth: d = h − cnom − Ø/2, and dv = d unless embedded items are present (4.3.3.2.1, 4.3.3.2.9). The control section lies at least dv/2 from the support edge.
- Design values: τcd = 0.3 · √fck / γc, 2.4.2.4 (3), and kg from Dmax, 4.3.3.2.1 (37).
- Strain of the bending bars εv at the level you choose (4.3.3.2.2), at level 2 εv = fyd/Es · md/mRd, equation (39), with mRd from a 0.85 · x stress block at fcd.
- Check 1, without shear reinforcement: vd ≤ vRd = kd · τcd · dv, 4.3.3.2.1 (35), with kd = 1/(1 + εv · d · kg), equation (36).
- Check 2, bending at the control section: md ≤ mRd.
- The slab as a web 1 m wide (4.3.3.3, 4.3.3.4), with z = 0.9 · d (4.3.3.4.2), αmin ≤ α ≤ 45°, 4.3.3.3.2 (42), and kc = 0.55 (4.2.1.7):
- Check 3, shear reinforcement: vd ≤ vRd,s = asw · 1000 · z · fyd · cotα, 4.3.3.4.3 (45).
- Check 4, compression field: vd ≤ vRd,c = 1000 · z · kc · fcd · sinα · cosα, 4.3.3.4.6 (47).
- Check 5, minimum shear reinforcement: ρw ≥ ρw,min = 0.001 · √(fck/30) · 500/fyk, 5.5.3.4 (116).
- Added tension in the bending bars, ftVd = vd · cotα/2, from 4.3.3.4.11 (51), to add to md/z.
What you enter and what you get
You enter the slab thickness, the face in tension, the cover and the bars in the direction of the shear: diameter, spacing, outer or inner layer, one or two layers, curtailment near the control section and the angle of the shear to the bars. Then embedded items, prestressing, the materials, the resistance factors, vd, a concentrated load near the support, the level for εv, md and nd, and the proposed links or studs with their spacings sl and st.
You get each check with its utilisation and clause. With no shear reinforcement chosen, the module shows the area needed when check 1 fails.
Assumptions and limits
- Level 1 takes the bars at yield, εv = fyd/Es (conservative). Level 3 uses the strain at 0.6 · d from the compression face, × 2.5. Bars that may yield take εv = 1.5 · fyd/Es (4.3.3.2.3).
- εv is increased by 50 % where bars are curtailed within d of the control section (4.3.3.2.4), and multiplied by 1/(sin⁴θ + cos⁴θ) when the shear runs at an angle to the bars (4.3.3.2.5).
- With prestress, level 2 uses md − mDd and mRd − mDd (4.3.3.2.6). A tension nd lowers mRd and raises the strain.
- A load within 2 · d of the support may have its share of vd reduced by a/(2 · d) (4.3.3.2.8). From an elastic analysis, vd may be averaged over 2 · d each side of its peak (4.3.3.1.5).
- With links, the concrete contribution of check 1 is not added.
Worked example
A 280 mm slab in fck = 30 N/mm² with Dmax = 32 mm, Ø14 bottom bars at 150 mm in the outer layer with 30 mm cover, under vd = 220 kN/m and md = 60 kNm/m at level 2, with Ø10 links at 200 × 200 mm; check 1, the shear resistance without shear reinforcement, does not pass, and the links carry vd. These are the module's own results for its worked example, computed when this page was built.
Inputs| Input | Value | Ref. |
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| Slab thickness h | 280 mm | |
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| Face in tension | Bottom face in tension (span) | |
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| Concrete cover at the tension face cnom | 30 mm | |
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| Position of these bars | Outer layer — next to the face | |
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| Bar diameter Ø | 14 mm | |
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| Bar spacing s | 150 mm | |
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| Layers in this direction | One layer | § 4.3.3.2.2 |
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| Bars curtailed near the control section | No — the bars run on past the control section | § 4.3.3.2.4 |
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| Direction of the shear θ | Along the bars | § 4.3.3.2.5 |
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| Conduits, pipes or inserts | None | § 4.3.3.2.9 |
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| Prestressing of the slab | None | § 4.3.3.2.6 |
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| Characteristic compressive strength fck | 30 N/mm² | § 3.1.2.2.2 |
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| Largest aggregate size Dmax | 32 mm | § 4.3.3.2.1 |
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| Characteristic yield strength of the bars fyk | 500 N/mm² | |
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| Modulus of elasticity Es | 205000 N/mm² | |
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| Resistance factor for concrete γc | 1.5 | § 2.4.2.6 |
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| Resistance factor for steel γs | 1.15 | § 2.4.2.6 |
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| Design shear force per metre vd | 220 kN/m | § 4.3.3.2.1 |
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| Concentrated load near the support | None | § 4.3.3.2.8 |
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| Level of approximation for εv εv | Level 2 — from md/mRd | § 4.3.3.2.2 |
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| Bending moment per metre at the control section md | 60 kNm/m | |
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| Duration factor ηt | 1.0 — both conditions of 4.2.1.3 met | § 4.2.1.3 |
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| Shear reinforcement | Vertical links or studs | § 4.3.3.3 |
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| Diameter of the elements Øsw | 10 mm | |
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| Spacing in the direction of the shear sl | 200 mm | |
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| Spacing across the shear st | 200 mm | |
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| Characteristic yield strength of the shear reinforcement fyk,w | 500 N/mm² | |
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| Condition of the slab αmin | Normal case — αmin = 30° | § 4.3.3.3.2 |
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Calculation: Effective depth| Step | Value | Ref. |
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| Depth of the barsh − cnom − Ø/2from the compression face | d = 243 mm | |
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| Bars per metreπ·Ø²/4 · 1000/s | as = 1,026 mm²/m | |
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| Effective depth for the sheardno embedded items | dv = 243 mm | § 4.3.3.2.1 |
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| Distance of the control section from the support edgeat least | dv/2 = 122 mm | § 4.3.3.2.1 |
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Calculation: Design values| Step | Value | Ref. |
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| Design yield strength of the barsfyk/γs | fyd = 434.8 N/mm² | § 2.4.2.5 (4) |
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| Design shear stress limit0.3·√fck/γc | τcd = 1.095 N/mm² | § 2.4.2.4 (3) |
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| Aggregate factor48/(16 + Dmax·min(1 ; (60/fck)²)) ≥ 1.2 | kg = 1.200 | § 4.3.3.2.1 (37) |
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Calculation: Design shear at the control section| Step | Value | Ref. |
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| Design shear for the checkvd | vd = 220.0 kN/m | |
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Calculation: Strain of the bending bars| Step | Value | Ref. |
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| Design compressive strength of the concreteηfc·ηt·fck/γc | fcd = 20.0 N/mm² | § 2.4.2.3 (2) |
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| Bending resistance per metreblock 0.85·x at fcd, bars at their design strengthabout mid-depth | mRd = 103.4 kNm/m | |
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| Strain of the bending barsfyd/Es·md/mRd | εv,0 = 1.230 ‰ | § 4.3.3.2.2 (39) |
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Calculation: Check 1 — shear without shear reinforcement: v_d ≤ v_Rd| Step | Value | Ref. |
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| Coefficient1/(1 + εv·d·kg), d in mm | kd = 0.736 | § 4.3.3.2.1 (36) |
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| Shear resistance per metrekd·τcd·dv | vRd = 195.9 kN/m | § 4.3.3.2.1 (35) |
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| Shear without shear reinforcementvd/vRd | η = 1.123 · Fail112 % of the capacity used | § 4.3.3.2.1 (35) |
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Calculation: Check 2 — bending at the control section: m_d ≤ m_Rd| Step | Value | Ref. |
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| Bending moment against the bending resistancemd/mRd | η = 0.580 · Pass58 % of the capacity used | |
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Calculation: The slab as a web, 1 m wide| Step | Value | Ref. |
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| Lever arm0.9·d | z = 219 mm | § 4.3.3.4.2 |
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| Design yield strength of the shear reinforcementfyk,w/γs | fyd,w = 434.8 N/mm² | § 2.4.2.5 (4) |
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| Design compressive strength of the concreteηfc·ηt·fck/γc | fcd = 20.0 N/mm² | § 2.4.2.3 (2) |
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| Smallest angle of the compression field | αmin = 30 ° | § 4.3.3.3.2 |
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| Reduction factor for the concrete strength | kc = 0.55 | § 4.2.1.7 |
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| Minimum shear reinforcement ratio0.001·√(fck/30)·500/fyk | ρw,min = 0.100 % | § 5.5.3.4 (116) |
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| Shear reinforcement per unit areaπ·Øsw²/4/(sl·st) | asw = 1,963 mm²/m² | |
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Calculation: Compression field| Step | Value | Ref. |
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| Angle of the compression fieldlargest resistance between αmin and 45° | α = 30.0 ° | § 4.3.3.3.2 (42) |
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Calculation: Check 3 — shear reinforcement: v_d ≤ v_Rd,s| Step | Value | Ref. |
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| Resistance of the shear reinforcementasw·1000·z·fyd·cotα | vRd,s = 323.4 kN/m | § 4.3.3.4.3 (45) |
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| Design shear against the shear reinforcementvd/vRd,s | η = 0.680 · Pass68 % of the capacity used | § 4.3.3.4.3 (45) |
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Calculation: Check 4 — compression field: v_d ≤ v_Rd,c| Step | Value | Ref. |
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| Resistance of the compression field1000·z·kc·fcd·sinα·cosα | vRd,c = 1,041.7 kN/m | § 4.3.3.4.6 (47) |
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| Design shear against the compression fieldvd/vRd,c | η = 0.211 · Pass21 % of the capacity used | § 4.3.3.4.6 (47) |
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Calculation: Check 5 — minimum shear reinforcement| Step | Value | Ref. |
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| Shear reinforcement ratioρw | η = 0.509 · Pass51 % of the capacity used | § 5.5.3.4 (116) |
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| Added tension in the bending barsvd·cotα/2half of (51), on the tension face: add it to md/z for these bars | ftVd = 190.5 kN/m | § 4.3.3.4.11 (51) |
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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 does the resistance depend on the bending moment?
The shear resistance falls as the bending bars stretch. At level 2 εv grows with md/mRd, which lowers kd and vRd.
What if the bars may yield over a support?
In a yield-line design, for example, take the plastic option: εv = 1.5 · fyd/Es (4.3.3.2.3).
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.
Related
Norm: SIA 262:2025 · Page updated: 29 October 2026