Weld Geometry — Clause 4.3

The effective throat thickness a of a fillet weld is the height of the largest triangle inscribed within the weld (measured perpendicular to the weld throat). For a full-penetration butt weld, a = t (plate thickness).

Leg Length s (mm) Throat a (mm) for 45° fillet Effective area per mm length
4 2.8 2.8 mm²/mm
5 3.5 3.5 mm²/mm
6 4.2 4.2 mm²/mm
8 5.7 5.7 mm²/mm
10 7.1 7.1 mm²/mm
12 8.5 8.5 mm²/mm

a = s / √2 = s × 0.707 for a standard 45° fillet. Deep penetration welds (with qualified WPS) may use a larger effective throat — up to s for certain processes.

Directional Method — Clause 4.5.3.2

The weld throat is subject to three stress components:

Design Criteria

σ_eq = √(σ⊥² + 3(τ⊥² + τ∥²)) ≤ f_u / (β_w × γ_M2)

And additionally: σ⊥ ≤ 0.9 × f_u / γ_M2

Where:

Correlation Factor β_w — Table 4.1

Steel Grade β_w
S235 0.80
S275 0.85
S355 0.90
S420 1.00
S460 1.00

The β_w factor accounts for the reduced ductility of higher-strength steels in the heat-affected zone. For S460, β_w = 1.00 fully offsets the strength increase — the weld design resistance per unit length for S460 is approximately the same as for S355 despite the higher parent material strength.

Simplified Method — Clause 4.5.3.3

The simplified method assumes the weld capacity is independent of the load direction:

F_w,Rd = f_vw,d × a

Where the design shear strength of the weld: f_vw,d = f_u / (√3 × β_w × γ_M2)

Weld Capacities per mm Run (Simplified Method)

Weld Throat a S235 (β_w = 0.80) S275 (β_w = 0.85) S355 (β_w = 0.90)
3 mm 0.80 kN/mm 0.72 kN/mm 0.82 kN/mm
4 mm 1.06 kN/mm 0.96 kN/mm 1.09 kN/mm
5 mm 1.33 kN/mm 1.20 kN/mm 1.37 kN/mm
6 mm 1.60 kN/mm 1.43 kN/mm 1.64 kN/mm
8 mm 2.13 kN/mm 1.91 kN/mm 2.19 kN/mm
10 mm 2.66 kN/mm 2.39 kN/mm 2.73 kN/mm

Worked Example 1 — 6 mm Fillet Weld, S355 Steel (Longitudinal)

Connection: 100 kN load transferred through 2 side fillet welds, 100 mm long each, 6 mm leg (4.2 mm throat), S355.

Parameter Value
Weld size 6 mm leg (a = 4.2 mm)
Steel S355 (f_u = 470 MPa)
β_w 0.90
Total weld length 2 × 100 = 200 mm

Directional Method

Stress component Value (kN/mm²)
τ∥ (parallel, assumed equal distribution) 100000 / (200 × 4.2) = 119 MPa
σ⊥, τ⊥ 0 (no transverse load)
σ_eq = √(0 + 3(0 + 119²)) 206 MPa
f_u / (β_w × γ_M2) 470 / (0.90 × 1.25) = 418 MPa
Check: 206 ≤ 418 OK

Simplified Method

Per mm capacity 1.64 kN/mm (from table)
Total capacity 200 × 1.64 = 328 kN
Applied load 100 kN
Utilization 100/328 = 0.30 OK

Worked Example 2 — 8 mm Transverse Fillet, S275 (End Fillet)

Problem: A 200 mm long transverse end fillet weld (8 mm leg, 5.7 mm throat) connects two S275 plates. Load = 450 kN applied perpendicular to the weld axis. Compare directional vs simplified methods.

Directional Method:

The load is perpendicular to the weld axis. Resolving into throat-plane components (45° fillet):

σ⊥ = (450,000 × sin 45°) / (200 × 5.7) = 450,000 × 0.707 / 1,140 = 279 MPa. τ⊥ = (450,000 × cos 45°) / (200 × 5.7) = 450,000 × 0.707 / 1,140 = 279 MPa. τ∥ = 0 (no longitudinal component).

σ_eq = √(279² + 3 × (279² + 0)) = √(77,841 + 3 × 77,841) = √(311,364) = 558 MPa. Resistance: f_u / (β_w × γ_M2) = 410 / (0.85 × 1.25) = 386 MPa.

558 MPa > 386 MPa — NOT OK. The 8 mm transverse weld is inadequate for 450 kN.

Revised: Increase to 10 mm leg (a = 7.1 mm). σ⊥ = τ⊥ = 450,000 × 0.707 / (200 × 7.1) = 224 MPa. σ_eq = √(224² + 3 × 224²) = √(200,704) = 448 MPa.

Still > 386 MPa. NOT OK with 10 mm.

Final revision: 12 mm leg (a = 8.5 mm) or increase weld length to 250 mm. With 10 mm leg and 250 mm length: σ⊥ = 450,000 × 0.707 / (250 × 7.1) = 179 MPa. σ_eq = √(4 × 179²) = 358 MPa < 386 MPa. OK.

Key lesson: Transverse welds have higher stress resultant (σ_eq = 2σ⊥ in the von Mises combination) than longitudinal welds. The directional method always shows transverse welds as weaker per unit length than the simplified method suggests — the simplified method is approximately 15% unconservative for purely transverse loading.

Butt Weld Design (Full Penetration)

Full penetration butt welds are designed as the parent metal: the weld strength equals the weaker part joined. No explicit weld check is required for correctly executed full-penetration butt welds with qualified WPS. Partial penetration butt welds are designed as deep fillet welds using the fillet weld method with the actual throat thickness.

Minimum and Maximum Weld Sizes

Per EN 1993-1-8 Clause 4.5:

For common thicknesses:

Thicker Part t (mm) Minimum a (mm) Minimum Leg s (mm)
≤ 4 2.5 3.5
4 < t ≤ 6 3.0 4.2
6 < t ≤ 12 4.0 5.7
12 < t ≤ 20 5.0 7.1
20 < t ≤ 30 6.0 8.5
> 30 8.0 11.3

The minimum weld size ensures adequate heat input to fuse the parent metal. Welds smaller than the minimum are prone to cracking from rapid cooling. For thin plates (t < 6 mm), the minimum leg size may exceed the plate thickness — in these cases, intermittent fillet welds (minimum 40 mm length at specified spacing) are permitted per EN 1993-1-8 Clause 4.3.2.

Frequently Asked Questions

What is the difference between the simplified and directional method for weld design? The simplified method (Clause 4.5.3.3) assumes the weld capacity is independent of load direction and is always conservative for longitudinal (side) fillets. The directional method (Clause 4.5.3.2) resolves stresses into σ⊥, τ⊥, and τ∥ components and can give higher capacity for longitudinal welds (up to 30% more) but lower capacity for purely transverse welds. The directional method is recommended for all heavily loaded connections because it correctly captures the direction-dependent strength of fillet welds.

What is the minimum fillet weld size per EN 1993-1-8? Per EN 1993-1-8 Clause 4.5, the minimum fillet weld throat thickness a is 2.5 mm for plates up to 4 mm thick, increasing to 8 mm for plates over 30 mm. The minimum leg length s = a × √2. For standard structural connections in S355 (8-20 mm plates), minimum a = 4-5 mm (leg s = 5.7-7.1 mm).

What is the correlation factor β_w and why does it increase with steel grade? β_w accounts for the reduced ductility and different metallurgy of the heat-affected zone (HAZ) in higher-strength steels. For S235 (β_w = 0.80), the weld metal overmatches the parent metal strength. For S460 (β_w = 1.00), the weld metal must match the parent metal strength while the HAZ has reduced toughness. The increasing β_w offsets the higher parent material f_u, so the design weld resistance f_vw,d is approximately constant across steel grades — the weld, not the parent metal, governs the design.

When should I use intermittent rather than continuous fillet welds? Intermittent fillet welds (minimum 40 mm long segments at specified spacing) are used for lightly loaded connections where the continuous weld strength far exceeds the applied load. They are common for stiffener-to-web connections in plate girders and for web-to-flange connections in built-up sections where shear flow is low. Maximum clear spacing = min(12t, 200 mm) for compression members and min(16t, 250 mm) for tension members. However, intermittent welds create corrosion traps and are discouraged for external applications and fatigue-loaded structures.

How does the directional method handle combined longitudinal and transverse loading? The directional method naturally handles combined loading by including both τ∥ (longitudinal) and σ⊥/τ⊥ (transverse) components in the von Mises equivalent stress. For a weld loaded at angle θ to its axis: σ_eq = √( (F sin θ / al)² × (sin²45° + 3cos²45°) + 3(F cos θ / al)² ). A weld loaded at 45° to its axis has σ_eq approximately 10% lower than a purely transverse weld, because part of the load is carried as longitudinal shear which is more efficiently resisted.

Related Pages

Educational reference only. Design per EN 1993-1-8:2005 Clause 4.5. β_w per Table 4.1. γ_M2 = 1.25. Verify electrode classification and weld procedure specification. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent verification.

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