Free Tool · EN 1992-1-1 §7.3.4

EN 1992-1-1 Crack Width Calculator

Crack width verification per EN 1992-1-1 §7.3.4. Computes σs under quasi-permanent loading via cracked-section analysis, then ρp,eff with hc,eff, maximum crack spacing sr,max (Eq 7.11), and wk = sr,max · (εsm − εcm). PASS/FAIL against Table 7.1N limit. C20/25–C50/60 · B500B · i18n EN/NL/DE.

NA x compression h_c,eff h d cracks w_k σ_c σ_s
wk = sr,max · (εsm − εcm) §7.3.4 sr,max = k3c + k1k2k4φ/ρp,eff Eq 7.11 εsm−εcm = [σs − ktfct,effp,eff·(1+αeρp,eff)] / Es Eq 7.9
Section & Loading Parameters
Diameter of the main longitudinal reinforcement bars.
Nominal cover to the main reinforcement. Used in s_r,max = k_3·c + k_1·k_2·k_4·φ/ρ_p,eff.
Centre-to-centre spacing of the main bars (c/c). For slabs this is typically the pitch of the reinforcement.
Characteristic yield strength of the longitudinal reinforcement. Typically 500 MPa (B500B).
EN 1992-1-1 Table 7.1N: XC1 → w_k,max = 0.4 mm; XC2/XC3/XC4/XD/XS → 0.3 mm.
Quasi-permanent bending moment (G + ψ₂·Q). For slabs enter moment per metre width.
Full breadth of the rectangular or T-beam web. For slabs enter 1000 mm for a 1 m strip.
Total depth of the cross-section from compression face to tension face.
Total area of tension reinforcement provided. For slabs enter mm²/m (use b = 1000 mm).
Crack Width Results
0.0730
wk (mm)
η = 24.3% of wk,max = 0.3 mm
PASS wk ≤ 0.3 mm (XC3)
Cracked Section
d (eff. depth)210.0 mm
α_e = E_s/E_cm6.09
E_cm32837 MPa
f_ctm2.90 MPa
Neutral axis x61.5 mm
I_cr358.8×10⁶ mm⁴
σ_s113.4 MPa
ρ_p,eff
h_c,eff62.8 mm
A_c,eff62821 mm²
ρ_p,eff0.03333
Crack Width
ε_sm − ε_cm3.580e-4 upper
s_r,max204.0 mm
w_k0.0730 mm
w_k,max0.3 mm
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Worked example — C30/37 slab, ϕ20@150, XC3

A 250 mm thick flat slab, b = 1000 mm strip, C30/37, ϕ20@150 (A_s = 2094 mm²/m), cover c = 30 mm, XC3 exposure, quasi-permanent moment M_qp = 45 kNm/m.

Inputs
φ = 20 mm, c = 30 mm, s = 150 mm, f_ck = 30 MPa, E_s = 200 000 MPa, M_qp = 45 kNm, b = 1000 mm, h = 250 mm, A_s = 2094 mm²
Material
f_ctm = 0.30×30^(2/3) = 2.896 MPa, E_cm = 22 000×(38/10)^0.3 = 32 837 MPa, α_e = 200 000/32 837 = 6.09
Geometry
d = 250 − 30 − 10 = 210 mm
Neutral axis
500x² + 6.09×2094·x − 6.09×2094×210 = 0 → x = 61.5 mm
I_cr
I_cr = 1000×61.5³/3 + 6.09×2094×(210−61.5)² = 358.8×10⁶ mm⁴
σ_s
σ_s = 45×10⁶×(210−61.5)×6.09/358.8×10⁶ = 113.4 MPa
h_c,eff
h_c,eff = min(2.5×(250−210), (250−61.5)/3, 125) = min(100, 62.8, 125) = 62.8 mm → A_c,eff = 62 800 mm²
ρ_p,eff
ρ_p,eff = 2094/62 800 = 0.0333
ε_sm−ε_cm
term = [113.4 − 0.4×(2.896/0.0333)×(1+6.09×0.0333)]/200 000 = 3.58×10⁻⁴; floor = 0.6×113.4/200 000 = 3.40×10⁻⁴ → upper governs
s_r,max
s_r,max = 3.4×30 + 0.8×0.5×0.425×20/0.0333 = 102 + 102 = 204 mm
w_k
w_k = 204×3.58×10⁻⁴ = 0.073 mm ≤ 0.30 mm → PASS
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Frequently asked questions

What is the quasi-permanent moment M_qp?

The quasi-permanent combination is M_qp = G·M_G + ψ₂·Q·M_Q per EN 1990 Eq.6.16b. ψ₂ is the quasi-permanent factor for variable loads (e.g. ψ₂ = 0.3 for imposed office loads). Use this moment — not the design ultimate moment — for crack width checks, which are serviceability verifications.

How is the neutral axis depth x calculated?

For a singly-reinforced cracked section: b/2·x² + α_e·A_s·x − α_e·A_s·d = 0. Solving this quadratic gives x, the neutral-axis depth above the tension reinforcement. α_e = E_s/E_cm is the modular ratio. The cracked second moment of area I_cr = b·x³/3 + α_e·A_s·(d−x)², used to compute σ_s = M_qp·(d−x)/I_cr·α_e.

What is ρ_p,eff and why does it matter?

ρ_p,eff = A_s / A_c,eff is the effective reinforcement ratio in the effective tension zone, where A_c,eff = b·h_c,eff with h_c,eff = min(2.5(h−d), (h−x)/3, h/2) per §7.3.2(3). It controls how effectively the reinforcement restrains crack growth — a higher ρ_p,eff means smaller crack widths.

What are the Table 7.1N limits for w_k,max?

EN 1992-1-1 Table 7.1N sets w_k,max based on exposure class: XC1 → 0.4 mm; XC2, XC3, XC4, XD1, XD2, XD3, XS1, XS2, XS3 → 0.3 mm (for reinforced concrete). For prestressed members or decompression requirements, lower limits apply — these are noted in the results panel.

What is the difference between k_t = 0.6 and k_t = 0.4?

k_t is a load-duration factor in the ε_sm − ε_cm formula (§7.3.4 Eq.7.9). k_t = 0.6 applies to short-term (instantaneous) loading; k_t = 0.4 applies to long-term loading (the quasi-permanent case used for crack width checks). The default in this calculator is k_t = 0.4.

When does the floor condition 0.6·σ_s/E_s govern?

When the reinforcement ratio is high (large ρ_p,eff), the concrete tension-stiffening term k_t·f_ct,eff/ρ_p,eff can become small, making the upper formula large. But if ρ_p,eff is low, the tension stiffening term can exceed σ_s and make the upper formula negative — in that case the floor 0.6·σ_s/E_s governs to prevent an unrealistically optimistic result.