Eurocode 2 Concrete Beam Shear Design Calculator

Professional Eurocode 2 shear design calculator for reinforced concrete beams. Compute VRd,c, VRd,max, and required shear reinforcement Asw/s per EN 1992-1-1 with clear, accessible UX.

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.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[V_{Rd,c} = \max\Big\{\big[C_{Rd,c}\, k\, (100\rho_l f_{ck})^{1/3} + k_1 \sigma_{cp}\big],\; \big[v_{min} + k_1 \sigma_{cp}\big]\Big\}\, b_w d\]
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
Formula (extracted LaTeX)
\[k = 1 + \sqrt{200/d} \le 2.0,\quad C_{Rd,c}=\frac{0.18}{\gamma_c},\quad k_1=0.15,\quad v_{min}=0.035\,k^{3/2}\sqrt{f_{ck}}\]
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}}
Formula (extracted LaTeX)
\[\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}\]
\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}
Formula (extracted LaTeX)
\[V_{Rd,s} = \frac{A_{sw}}{s}\, z\, f_{ywd}\, \cot\theta,\quad z \approx 0.9 d,\quad f_{ywd}=\min\!\left(\frac{f_{yk}}{\gamma_s},\; 435\text{ MPa}\right)\]
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)
Formula (extracted LaTeX)
\[V_{Rd,\max} = \frac{\alpha_{cw}\, b_w\, z\, v_1\, f_{cd}}{\cot\theta + \tan\theta},\quad v_1 = 0.6\left(1-\frac{f_{ck}}{250}\right),\quad \alpha_{cw}\approx 1.0\]
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
Formula (extracted LaTeX)
\[\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]\]
\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]
Formula (extracted text)
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): $\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

Full original guide (expanded)

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.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[V_{Rd,c} = \max\Big\{\big[C_{Rd,c}\, k\, (100\rho_l f_{ck})^{1/3} + k_1 \sigma_{cp}\big],\; \big[v_{min} + k_1 \sigma_{cp}\big]\Big\}\, b_w d\]
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
Formula (extracted LaTeX)
\[k = 1 + \sqrt{200/d} \le 2.0,\quad C_{Rd,c}=\frac{0.18}{\gamma_c},\quad k_1=0.15,\quad v_{min}=0.035\,k^{3/2}\sqrt{f_{ck}}\]
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}}
Formula (extracted LaTeX)
\[\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}\]
\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}
Formula (extracted LaTeX)
\[V_{Rd,s} = \frac{A_{sw}}{s}\, z\, f_{ywd}\, \cot\theta,\quad z \approx 0.9 d,\quad f_{ywd}=\min\!\left(\frac{f_{yk}}{\gamma_s},\; 435\text{ MPa}\right)\]
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)
Formula (extracted LaTeX)
\[V_{Rd,\max} = \frac{\alpha_{cw}\, b_w\, z\, v_1\, f_{cd}}{\cot\theta + \tan\theta},\quad v_1 = 0.6\left(1-\frac{f_{ck}}{250}\right),\quad \alpha_{cw}\approx 1.0\]
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
Formula (extracted LaTeX)
\[\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]\]
\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]
Formula (extracted text)
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): $\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

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.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[V_{Rd,c} = \max\Big\{\big[C_{Rd,c}\, k\, (100\rho_l f_{ck})^{1/3} + k_1 \sigma_{cp}\big],\; \big[v_{min} + k_1 \sigma_{cp}\big]\Big\}\, b_w d\]
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
Formula (extracted LaTeX)
\[k = 1 + \sqrt{200/d} \le 2.0,\quad C_{Rd,c}=\frac{0.18}{\gamma_c},\quad k_1=0.15,\quad v_{min}=0.035\,k^{3/2}\sqrt{f_{ck}}\]
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}}
Formula (extracted LaTeX)
\[\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}\]
\sigma_{cp} = \min\!\left(\frac{N_{Ed}}{A_c},\; 0.2\,f_{cd}\right),\quad f_{cd}=\alpha_{cc}\frac{f_{ck}}{\gamma_c}
Formula (extracted LaTeX)
\[V_{Rd,s} = \frac{A_{sw}}{s}\, z\, f_{ywd}\, \cot\theta,\quad z \approx 0.9 d,\quad f_{ywd}=\min\!\left(\frac{f_{yk}}{\gamma_s},\; 435\text{ MPa}\right)\]
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)
Formula (extracted LaTeX)
\[V_{Rd,\max} = \frac{\alpha_{cw}\, b_w\, z\, v_1\, f_{cd}}{\cot\theta + \tan\theta},\quad v_1 = 0.6\left(1-\frac{f_{ck}}{250}\right),\quad \alpha_{cw}\approx 1.0\]
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
Formula (extracted LaTeX)
\[\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]\]
\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]
Formula (extracted text)
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): $\frac{A_{sw}}{s} \ge \frac{0.08\, \sqrt{f_{ck}}}{f_{yk}}\, b_w\quad \big[\text{units: } \mathrm{mm}^2/\mathrm{mm}\big]$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
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Formulas

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Version 0.1.0-draft
Citations

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Changelog
  • 0.1.0-draft — 2026-01-19: Initial draft (review required).