Free Tool · EN 1992-1-1 §6.2

EN 1992-1-1 RC Beam Shear Design Calculator

RC beam shear design per EN 1992-1-1 §6.2. Computes VRd,c without shear reinforcement (§6.2.2), then VRd,s + VRd,max with vertical or inclined stirrups using variable strut inclination (§6.2.3). Auto-selects the governing resistance and shows utilisation η = VEd/VRd. C20/25–C50/60 · B500B · γ_C = 1.50 · i18n EN/NL/DE.

V_Ed b_w A_sl tension ∅... θ concrete strut cot θ = 1.0–2.5
VRd,c = [CRd,c·k·(100·ρl·fck)1/3]·bw·d §6.2.2 | VRd,s = (Asw/s)·z·fywd·cot θ §6.2.3 | VRd,max = αcw·bw·z·ν1·fcd/(cot θ+tan θ) eq.6.9
Beam & Loading Parameters
Cross-section & Loading
Web breadth of the T- or rectangular section. For a rectangular beam b_w = b.
Effective depth to the centroid of the tensile reinforcement. Typically d ≈ h − c − φ_link − φ_long/2.
Design shear force at the check section (e.g. at distance d from the support face).
Characteristic yield strength of the transverse reinforcement (stirrups). Default 500 MPa (B500B).
Area of the longitudinal tension reinforcement (bottom face). Required for ρ_l in V_Rd,c.
Optional axial force (compression positive). Used to compute σ_cp in V_Rd,c. Leave at 0 for simply-supported beams.
Stirrup input mode
"Design mode" computes the required A_sw/s and spacing from V_Ed. "Check mode" validates a provided stirrup configuration.
Area of one stirrup leg (both legs of a closed link). E.g. Ø8 two-leg: A_sw = 2×50.3 = 100.6 mm².
Centre-to-centre spacing of the stirrups along the beam axis.
Number of active stirrup legs crossing the potential crack. Typically 2 for a closed link, 4 for a疯子 cage.
cot θ (strut inclination)
Cotangent of the concrete strut angle θ. Range 1.0 (θ=45°) to 2.5 (θ=21.8°). EN §6.2.3(2) Table NA.1. Lower cot θ = steeper strut = lower V_Rd,s but higher V_Rd,max.
Shear Resistance Results
175.18 kN
V_Rd governing
Shear reinf. req. cot θ = 2
103% η = V_Ed / V_Rd
FAIL
§6.2.2 V_Rd,c (no shear reinf.)
k = 1+√(200/d) 1.67
ρ_l = A_sl/(b_w·d) ≤ 0.02 0.009
C_Rd,c 0.12
v_min (eq.6.3N) 0.414 MPa
V_Rd,c (eq.6.2a) 80.25 kN
V_Rd,c,min floor 55.25 kN
§6.2.3 Shear reinforcement
cot θ / θ 2 / 26.6°
V_Rd,max (strut crushing) 431.39 kN
V_Rd,s (stirrups) 175.18 kN
A_sw,min/s (§9.2.2 eq.9.5N) 0.263 mm²/mm
A_sw/s required 0.517 mm²/mm
s_l,max (§9.2.2 eq.9.6N) 333.8 mm
Step 1 k = 1 + √(200/445) = —
Step 2 ρ_l = —/(—×—) = —
Step 3 V_Rd,c = [—×—×(100×—×—)1/3] × — × — = — kN
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Worked example — simply-supported beam
A 300 × 500 mm simply-supported beam carries V_Ed = 180 kN at d = 445 mm from the support. Concrete C30/37, B500B stirrups Ø8@200 two-legs (A_sw = 100.6 mm²), no axial force.
Input
b_w = 300 mm, d = 445 mm, f_ck = 30 MPa, f_yk = 500 MPa, V_Ed = 180 kN, A_sl = 1200 mm²
§6.2.2 V_Rd,c
k = 1+√(200/445) = 1.67, ρ_l = 1200/(300×445) = 0.0090, C_Rd,c = 0.12
V_Rd,c
V_Rd,c = (0.12×1.67×(100×0.009×30)^(1/3)) × 300 × 445 = 91.4 kN — minimum governs
Result
V_Ed (180 kN) > V_Rd,c (91.4 kN) → shear reinforcement required
§6.2.3 V_Rd,s
cot θ = 2.0 (default), z = 0.9×445 = 400.5 mm, f_ywd = 435 MPa
V_Rd,s
V_Rd,s = (100.6/200) × 400.5 × 435 × 2.0 = 175.5 kN
V_Rd,max
V_Rd,max = (300×400.5×0.54×17) / (2.0+0.5) = 367.6 kN
Governing
min(175.5, 367.6) = 175.5 kN — PASS η = 180/175.5 = 102.6% → increase stirrup area
Automate RC beam shear design for every member
FrameAI Pro extracts all RC beams from your structural drawing PDF and runs full EN 1992-1-1 §6.2 shear checks — V_Rd,c, V_Rd,s, V_Rd,max, minimum stirrups §9.2.2 — for every member automatically.
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Related tools
Punching shear (§6.4) RC Column (§5.8.8) Retaining wall (EN 1992-1-1 + EN 1997-1) Civil Engineering overview

Frequently asked questions

When is shear reinforcement required?

Per EN 1992-1-1 §6.2.2, shear reinforcement is required when V_Ed > V_Rd,c. V_Rd,c = [C_Rd,c·k·(100·ρ_l·fck)^(1/3) + k_1·σ_cp]·b_w·d with a minimum floor V_Rd,c,min = (v_min + k_1·σ_cp)·b_w·d. The larger of the two governs. If V_Ed ≤ V_Rd,c, the concrete alone carries the shear — but minimum stirrups §9.2.2 still apply.

What is the variable strut inclination method (§6.2.3)?

The variable strut inclination method (§6.2.3) models the concrete compression field as a series of struts at angle θ to the beam axis. The designer chooses cot θ in the range [1.0, 2.5]. A lower cot θ (steeper strut = larger θ) gives a higher V_Rd,max but requires more stirrup steel (V_Rd,s ∝ cot θ). The calculator optimises cot θ automatically to the largest value that still satisfies V_Rd,max ≥ V_Ed — the most economical design.

What is the difference between V_Rd,s and V_Rd,max?

V_Rd,s is the shear resistance of the stirrup reinforcement, computed from A_sw/s · z · f_ywd · cot θ (eq.6.8). V_Rd,max is the concrete strut crushing resistance, computed from α_cw · b_w · z · ν_1 · f_cd / (cot θ + tan θ) (eq.6.9). The actual shear resistance V_Rd = min(V_Rd,s, V_Rd,max) — whichever is smaller governs. If V_Rd,s < V_Rd,max the stirrups yield first (steel governs). If V_Rd,max < V_Rd,s the concrete crushes (concrete governs).

How do I choose the stirrup area A_sw?

A_sw is the total cross-sectional area of one stirrup leg (both legs of a closed link count together). E.g. Ø8 closed link: A_sw = 2 × (π×8²/4) = 2 × 50.3 = 100.6 mm². Ø10 closed link: A_sw = 2 × 78.5 = 157.0 mm². In design mode, the calculator computes the required A_sw/s from V_Ed and suggests a spacing s that fits your chosen stirrup size.

What is the minimum shear reinforcement (§9.2.2)?

Per EN 1992-1-1 §9.2.2, A_sw,min/s = 0.08·√f_ck·b_w/f_yk (eq.9.5N). This is a mandatory floor regardless of shear demand. The calculator enforces this in both design and check modes. If your provided A_sw/s falls below this floor, the calculator flags a warning and uses the minimum value.

How does the German National Annex differ?

The German National Annex (DIN EN 1992-1-1/NA) uses a lower C_Rd,c coefficient: 0.15/γ_C instead of 0.18/γ_C. This reduces V_Rd,c by about 17% compared to the EN recommended value, making shear reinforcement more likely to be required. Select "DE" in the National Annex toggle to apply this.