Weld calculation per EN 1993-1-8
Verify fillet welds per EN 1993-1-8: load and weld geometry yield the stress components in the effective throat plane and the utilisation per the directional and the simplified method – including the required minimum throat thickness.
Fillet weld calculator (EN 1993-1-8)
Tension perpendicular to the plate plane: decomposition into σ⊥ = τ⊥ with the factor 1/√2.
corresponds to leg length z ≈ 5.7 mm
Adjust material values (f_u, β_w)
f_u of the weaker part, not of the filler metal. S355 depending on source 470 (N/M grades), 490 (EN 10025-2, default) or 510 N/mm² (EN 1993-1-1 Tab. 3.1).
Defaults per the German National Annex: S235 0.80 / S275 0.85 / S355 0.90 / S420 0.88 / S460 0.85. EN Table 4.1 gives 1.0 for S420/S460.
Stress components in the throat plane
- Effective length l_eff
- 200 mm / A_w = 1,600 mm²
- σ⊥
- 132.6 N/mm²
- τ⊥
- 132.6 N/mm²
- τ∥
- 0 N/mm²
Directional method
η = 0.74- σ_wv ≤ f_u/(β_w·γ_M2)
- 265.2 ≤ 360 N/mm²
- η₁
- 0.737
- σ⊥ ≤ 0.9·f_u/γ_M2
- 132.6 ≤ 259.2 N/mm²
- η₂
- 0.512
Verification satisfied.
Simplified method
η = 0.9- f_vw,d
- 207.8 N/mm²
- F_w,Ed ≤ F_w,Rd
- 750 ≤ 831.4 N/mm
Direction-independent and about 22 % more conservative for transverse tension; identical to the directional method for pure longitudinal shear.
Static resistance verification for predominantly static loading per DIN EN 1993-1-8:2010-12 with the German National Annex (model angle 90°, S235 to S460, t ≤ 40 mm). No fatigue verification per EN 1993-1-9.
Formulas and fundamentals
The calculator performs the static resistance verification of fillet welds per EN 1993-1-8, section 4. The effective throat area A_w = n · a · l_eff is notionally rotated into the root line; the stresses are assumed to be uniformly distributed. The effective length is l_eff = l − 2 · a unless the weld is full-size over its whole length including the ends. The throat thickness a is the height of the inscribed triangle – not the leg length z (for 90°: a = z / √2).
Because the throat plane of a right-angle joint is inclined at 45° to the load direction, a nominal stress transverse to the weld splits equally into a normal and a shear component: σ⊥ = τ⊥ = σ_w / √2. Forces parallel to the weld axis produce pure longitudinal shear τ∥. For the tee joint in bending, the weld group is treated as a line section: W_w = a · l_eff² / 3 for the double fillet weld, plus τ∥ = V_Ed / (2 · a · l_eff) from the shear force. The calculator conservatively superimposes the bending maximum at the weld end with the shear maximum.
Directional method: both conditions must be satisfied. First the comparison value √(σ⊥² + 3 · (τ⊥² + τ∥²)) ≤ f_u / (β_w · γ_M2), second the additional condition σ⊥ ≤ 0.9 · f_u / γ_M2. Here f_u is the tensile strength of the weaker of the connected parts and β_w the correlation factor of the steel grade (S235: 0.80, S275: 0.85, S355: 0.90; per the German National Annex S420: 0.88 and S460: 0.85).
Simplified method: the resultant force per unit length of weld F_w,Ed is compared, independent of its direction, against F_w,Rd = a · f_vw,d, with the design shear strength of the weld f_vw,d = f_u / (√3 · β_w · γ_M2). For pure longitudinal shear both methods give the same result; for tension transverse to the weld the simplified method is more conservative by the factor √3/√2 ≈ 1.22. The calculator shows both utilisations side by side.
In addition the calculator checks the detailing rules: absolute minimum throat a ≥ 3 mm, thickness-dependent guide value a ≥ √(max t) − 0.5 per the German National Annex (at least 5 mm for t ≥ 30 mm), maximum thickness a ≤ 0.7 · min t as a practice rule, minimum load-bearing length l_eff ≥ max(30 mm; 6 · a), and the long-weld note for lap joints with L_j > 150 · a. The required throat thickness a_req is determined iteratively from the utilisation.
Worked example
Given: tee joint with a double fillet weld, plate 200 × 12 mm on a base plate t = 20 mm, tensile force N_Ed = 300 kN perpendicular to the plate plane. Steel S235 (f_u = 360 N/mm², β_w = 0.80), γ_M2 = 1.25, throat a = 4 mm, weld length l = 200 mm with full-size ends, so l_eff = 200 mm and A_w = 2 · 4 · 200 = 1600 mm².
Directional method: the nominal stress σ_w = 300,000 / 1600 = 187.5 N/mm² splits into σ⊥ = τ⊥ = 187.5 / √2 = 132.6 N/mm². Comparison value √(132.6² + 3 · 132.6²) = 265.2 N/mm² ≤ 360 N/mm², hence η₁ = 0.74. Additional condition: 132.6 ≤ 0.9 · 360 / 1.25 = 259.2 N/mm², η₂ = 0.51 – the verification is satisfied, condition 1 governs.
Simplified method: F_w,Ed = 300,000 / (2 · 200) = 750 N/mm against F_w,Rd = 207.8 · 4 = 831.4 N/mm gives η = 0.90. This shows the conservatism of the simplified method for transverse tension (0.90 instead of 0.74). Detailing checks: a_min = √20 − 0.5 = 3.97 mm ≤ 4 mm and a_max = 0.7 · 12 = 8.4 mm are satisfied.
Frequently asked questions
What is the difference between throat thickness a and leg length z?
The throat thickness a is the height of the triangle inscribed in the weld cross-section down to the theoretical root point. The leg length z often shown on drawings is the cathetus of the weld triangle; for a right-angle joint a = z / √2. Mixing them up distorts the verification by a good 40 percent.
Which method should I use – directional or simplified?
The directional method is more economical because it accounts for the direction of loading. The simplified method is about 22 percent more conservative for tension transverse to the weld and identical for pure longitudinal shear. Both are code-compliant; the calculator shows both utilisations side by side.
Which tensile strength f_u applies?
That of the weaker of the connected parts, never that of the filler metal. For S355 the sources quote 470, 490 or 510 N/mm²: the product standard EN 10025-2 gives 490 N/mm² (calculator default), EN 1993-1-1 Table 3.1 recommends 510 N/mm², and N/NL and M/ML grades are at 470 N/mm². The value can be overridden.
Why does β_w for S420 and S460 differ from the EN table?
EN 1993-1-8 Table 4.1 gives 1.0 for both grades. The German National Annex instead uses 0.88 (S420) and 0.85 (S460) together with γ_M2 = 1.25. The calculator uses the German values as defaults; the EN combination can be set via the overrides.
When do I have to reduce the effective length by 2 · a?
Whenever the start and end of the weld are not executed at full size (end craters). If the weld is full-size including its ends – for example by welding all around or using run-off tabs – the full length may be used. Welds shorter than max(30 mm; 6 · a) are considered not load-bearing.
Does the calculator also cover dynamically loaded welds?
No. The verification applies to predominantly static loading. Under fluctuating loads the fatigue verification per EN 1993-1-9 with detail categories and S-N curves is additionally required, which this tool deliberately does not cover.