This authoritative tool helps structural engineers and civil engineering students design the shear resistance of reinforced concrete beams according to Eurocode 2 (EN 1992-1-1). It computes concrete shear resistance VRd,c, checks VRd,max, and determines the required shear reinforcement Asw/s and spacing for selected stirrups, ensuring clarity, accessibility, and traceability to the standard.

Data Source and Methodology

Authoritative standard: EN 1992-1-1:2004 + A1:2014 Eurocode 2 — Design of concrete structures — Part 1-1: General rules and rules for buildings. Clauses 6.2.2 (VRd,c), 6.2.3 (VRd,max and VRd,s), and 9.2.2 (minimum shear reinforcement). Direct reference (open access where available):

All calculations are strictly based on the formulas and data provided by this source.

The Formula Explained

Concrete shear resistance (EN 1992-1-1, 6.2.2):

$$V_{Rd,c} = \max\Big\{\big[C_{Rd,c}\, k\, (100\rho_l f_{ck})^{1/3} + k_1 \sigma_{cp}\big],\; \big[v_{min} + k_1 \sigma_{cp}\big]\Big\}\, b_w d$$

with

$$k = 1 + \sqrt{200/d} \le 2.0,\quad C_{Rd,c}=\frac{0.18}{\gamma_c},\quad k_1=0.15,\quad v_{min}=0.035\,k^{3/2}\sqrt{f_{ck}}$$

Axial stress (compression positive):

$$\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}$$

Shear with reinforcement (EN 1992-1-1, 6.2.3):

$$V_{Rd,s} = \frac{A_{sw}}{s}\, z\, f_{ywd}\, \cot\theta,\quad z \approx 0.9 d,\quad f_{ywd}=\min\!\left(\frac{f_{yk}}{\gamma_s},\; 435\text{ MPa}\right)$$

Maximum shear capacity:

$$V_{Rd,\max} = \frac{\alpha_{cw}\, b_w\, z\, v_1\, f_{cd}}{\cot\theta + \tan\theta},\quad v_1 = 0.6\left(1-\frac{f_{ck}}{250}\right),\quad \alpha_{cw}\approx 1.0$$

Minimum shear reinforcement (EN 1992-1-1, 9.2.2):

$$\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]$$

Glossary of Variables

  • b_w: web width (mm); d: effective depth (mm); h: overall depth (mm)
  • f_ck: characteristic concrete strength (MPa); f_cd: design concrete strength (MPa)
  • f_yk: characteristic yield strength of stirrups (MPa); f_ywd: design yield strength (MPa)
  • γ_c, γ_s: partial safety factors for concrete and steel; α_cc: coefficient for long-term effects
  • ρ_l: longitudinal tension reinforcement ratio A_sl/(b_w·d) (use % as input; capped at 2% in the formula)
  • N_Ed: axial force at ULS (kN), compression positive; σ_cp: axial stress (MPa)
  • V_Ed: applied design shear (kN); V_Rd,c: concrete shear resistance (kN)
  • V_Rd,max: maximum shear capacity with truss model (kN)
  • A_sw: total stirrup steel area per set (mm²); s: stirrup spacing (mm)
  • θ: angle of compression strut; z ≈ 0.9d; v_1 = 0.6(1 − f_ck/250)

Worked Example

How It Works: A Step-by-Step Example

Given: b_w = 300 mm, d = 500 mm, h = 600 mm; f_ck = 30 MPa; f_yk = 500 MPa; γ_c = 1.5; γ_s = 1.15; α_cc = 1.0; ρ_l = 1.0%; V_Ed = 250 kN; N_Ed = 0; θ = 45°.

  1. Compute k = 1 + √(200/d) = 1 + √(200/500) = 1.632 ≤ 2.0.
  2. f_cd = α_cc f_ck / γ_c = 1.0·30/1.5 = 20 MPa. σ_cp = 0.
  3. C_Rd,c = 0.18/γ_c = 0.12. v_min = 0.035 k^(3/2) √f_ck ≈ 0.035·(1.632)^{1.5}·√30 ≈ 0.35 MPa.
  4. v_Rd,c = max[0.12·1.632·(100·0.01·30)^{1/3}, v_min] ≈ max[0.12·1.632·(30)^{1/3}, 0.35] ≈ 0.56 MPa.
  5. V_Rd,c = v_Rd,c·b_w·d = 0.56·300·500 / 1000 ≈ 84 kN.
  6. Since V_Ed > V_Rd,c, shear reinforcement is required. z ≈ 0.9d = 450 mm; cotθ = 1.
  7. f_ywd = min(f_yk/γ_s, 435) = min(500/1.15, 435) = 435 MPa. Required Asw/s = (250−84)·1000 / (450·435·1) ≈ 0.85 mm²/mm = 850 mm²/m.
  8. Minimum Asw/s = 0.08 √f_ck b_w / f_yk = 0.08·√30·300/500 ≈ 0.26 mm²/mm = 260 mm²/m. Governing is 850 mm²/m.
  9. Choose φ8 stirrups with 2 legs: A_sw = 2 · π·8²/4 = 100.5 mm². Spacing s = A_sw / (Asw/s) = 100.5 / 0.85 ≈ 118 mm. Check s ≤ min(0.75d, 600) ⇒ 0.75d = 375 mm, so OK.
  10. Check V_Rd,max with v_1 = 0.6(1−30/250)=0.528: V_Rd,max = α_cw b_w z v_1 f_cd /(cotθ+tanθ) = 1·300·450·0.528·20/(1+1)/1000 ≈ 712 kN. Since 250 kN ≤ 712 kN, OK.

The calculator reproduces these steps automatically and provides pass/fail tags and spacing recommendations.

Frequently Asked Questions (FAQ)

Do I need to model z precisely?

For beams with typical reinforcement layouts, z ≈ 0.9d is acceptable. If you have a detailed section analysis, you may replace z accordingly.

What if V_Ed exceeds V_Rd,max?

The section is not adequate in shear, regardless of stirrups. Increase b_w, d, f_ck, reduce θ (increase cotθ), or revise the structural system.

Can I input tension (negative N_Ed)?

Eurocode’s σ_cp term benefits compression only. Tensile axial force is conservatively ignored (σ_cp is not taken as negative here).

How is minimum stirrup spacing checked?

The tool limits spacing to s ≤ min(0.75·d, user limit). National Annexes may impose additional limits; verify against your NA.

Does the tool consider shear span-to-depth (a/d)?

No. The calculator focuses on sectional shear checks. For very small a/d (deep beams), specific provisions apply and are outside this tool’s scope.

Which units should I use?

Input geometry in mm, strengths in MPa, forces in kN. The tool handles conversions consistently.

Tool developed by Ugo Candido.
Content verified by Structural Engineering Reviewer.
Last reviewed for accuracy on: .