-------- | ---------- | ---- | ---------------------------------- | | S235 | 360 | 0.80 | Base steel | | S275 | 410 | 0.85 | Common for plates, secondary steel | | S355 | 470 | 0.90 | Standard structural sections | | S420 | 520 | 1.00 | Higher strength sections | | S460 | 540 | 1.00 | Most onerous ÃÂòw factor |
Weld Design Strengths (Simplified Method)
| Steel Grade | fu (N/mmÃÂò) | fvw,d (N/mmÃÂò) | Capacity per mm throat (N/mm) |
|---|---|---|---|
| S235 | 360 | 208 | 208 |
| S275 | 410 | 223 | 223 |
| S355 | 470 | 241 | 241 |
| S460 | 540 | 249 | 249 |
Note: fvw,d = fu / (âÃÂÃÂ3 ÃÂàÃÂòw ÃÂàÃÂóM2). For S355 into S275: fu = 410 MPa (weaker part).
Minimum and Maximum Weld Sizes
| Thickness of Thicker Part t (mm) | Minimum Throat a (mm) | Minimum Leg Length (mm) | Maximum Weld Size |
|---|---|---|---|
| t âÃÂä 10 | 3 | 4 | 0.7t (butt) |
| 10 < t âÃÂä 20 | 4 | 6 | 0.7t |
| 20 < t âÃÂä 30 | 5 | 7 | 0.7t |
| 30 < t âÃÂä 50 | 6 | 8 | 0.7t |
| t > 50 | 8 | 11 | 0.7t (or as specified) |
Minimum weld length = 6a or 40 mm (whichever greater).
Effective Weld Length
The effective length of a fillet weld: [ L*{eff} = L*{actual} - 2a ]
For a 200 mm long fillet weld with 6 mm throat: Leff = 200 - 12 = 188 mm.
For end-loaded welds (long joints), the effective length is reduced if L > 150a:
- Reduction factor ÃÂòLw = 1.2 - 0.2 ÃÂà(Lw / 150a) but not < 0.6 or > 1.0
Worked Example — Fillet Welded Bracket
Given:
- Bracket: 10 mm plate in S275 welded to column flange
- Design shear: 150 kN (in-plane vertical)
- Fillet weld: 6 mm leg length (a = 4.2 mm), S355 member
- Weld length: 2 ÃÂÃÂ 150 mm vertical (return welds at top)
Step 1 — Determine design strength: Weaker part = S275: fu = 410 MPa, ÃÂòw = 0.85) fvw,d = 410 / (âÃÂÃÂ3 ÃÂà0.85 ÃÂà1.25) = 222.8 N/mmÃÂò
Step 2 — Weld capacity per unit length: Fw,Rd = 222.8 ÃÂÃÂ 4.2 = 935.8 N/mm
Step 3 — Total weld capacity: Total effective length = 2 ÃÂà(150 - 2 ÃÂà4.2) = 283.2 mm Total capacity = 283.2 ÃÂà935.8 ÃÂà10âÃÂûÃÂó = 265.0 kN
Step 4 — Check: UT = 150 / 265.0 = 0.566 — Satisfactory
Step 5 — Check weld return (top): The 12 mm return at top (typical detail) adds ~2% capacity, conservatively ignored.
Worked Example — Full Penetration Butt Weld
Given:
- 20 mm plate splice in S355
- Tension force: 1500 kN
- Plate width: 300 mm
Full penetration butt weld:
- Design strength = fy / ÃÂóM0 = 355 / 1.0 = 355 N/mmÃÂò (tension yield)
- OR fu / ÃÂóM2 = 470 / 1.25 = 376 N/mmÃÂò (tension fracture on net section)
- Conservative: use fy/ÃÂóM0 = 355 N/mmÃÂò
Capacity = 300 ÃÂà20 ÃÂà355 / 1.0 ÃÂà10âÃÂûÃÂó = 2130 kN > 1500 kN — Satisfactory
Full penetration butt welds (subject to NDT verification) are designed as equal to the parent metal strength. No throat thickness calculation is required.
Partial Penetration Butt Welds
| Design Method | Clause | Commentary |
|---|---|---|
| Treated as fillet weld | BS EN 1993-1-8 Clause 4.1(2) | Use effective throat = depth of penetration |
| Ultimate limit state | — | Design for fu/ÃÂóM2 in tension |
| Fatigue detail | BS EN 1993-1-9 | Treat as Class 36 or 40 depending on NDT |
Weld Electrode Selection for UK Practice
| Steel Grade | Recommended Electrode (MAG) | Electrode Classification (EN ISO 14341) |
|---|---|---|
| S235 | G 42 4 M21 3Si1 | Yield âÃÂÃÂ¥ 420 MPa, Charpy 47J at 0ÃÂðC |
| S275 | G 42 4 M21 3Si1 | Same, adequate for S275 |
| S355 | G 46 4 M21 3Si1 | Yield âÃÂÃÂ¥ 460 MPa |
| S420 | G 50 4 M21 3Si1 | Yield âÃÂÃÂ¥ 500 MPa |
| S460 | G 55 4 M21 3Si1 | Yield âÃÂÃÂ¥ 550 MPa |
For manual metal arc (MMA): E42 (S275), E46 (S355) per EN ISO 2560.
Design Resources
- UK Connection Design — General connection guidance
- UK Steel Properties — Material properties
- UK HSS Connections — Hollow section joints
- UK Steel Beam Sizes — Section data
- UK Bolt Capacity — Alternative connection methods
- All UK References
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Frequently Asked Questions
How is fillet weld capacity calculated per BS EN 1993-1-8?
Two methods: simplified (Fw,Rd = fvw,d ÃÂàa per unit length, fvw,d = fu/âÃÂÃÂ3 / (ÃÂòw ÃÂàÃÂóM2)) and directional (stresses resolved on throat area). The simplified method is conservative and preferred for UK office use. UK NA specifies ÃÂòw = 0.85 for S275, 0.90 for S355, ÃÂóM2 = 1.25. For a 6 mm fillet weld on S355: fvw,d = 470/(âÃÂÃÂ3 ÃÂà0.9 ÃÂà1.25) = 241 N/mmÃÂò, capacity = 241 ÃÂà4.2 = 1012 N/mm.
What is the minimum fillet weld size in UK practice?
Minimum fillet weld throat thickness a = 3 mm per BS EN 1090-2. For thicker plates (> 20 mm), a minimum of 4-5 mm is typically specified. Minimum weld length = 40 mm or 6a (whichever greater). Maximum weld leg length should not exceed 0.7 ÃÂÃÂ plate thickness (to avoid excessive heat input). The actual weld size is determined by the design force — UK fabricators prefer 6-8 mm leg length (4.2-5.6 mm throat) as the economic range for workshop welding.
When should the directional method be used instead of the simplified method?
The simplified method (Clause 4.5.3.2) is adequate for most UK design office applications and is more conservative. The directional (resolved stress) method (Clause 4.5.3.3) should be used when: (a) the weld is subject to complex multi-axial stress, (b) the weld orientation relative to the load direction significantly affects capacity (a transversely loaded fillet weld has ~25% higher capacity than a longitudinally loaded weld), or (c) the connection is highly stressed and the simplified method is too conservative.
What is the difference between full and partial penetration butt welds?
Full penetration butt welds (Clause 4.7) develop the full strength of the parent metal and do not require capacity checks beyond NDT verification. Partial penetration butt welds (Clause 4.2) are designed as fillet welds using the effective throat (depth of penetration). Full penetration is required for: tension members, moment connections, and fatigue-loaded details. Partial penetration is acceptable for: compression splices, shear connections, and non-structural attachments.
How are long fillet welds designed per BS EN 1993-1-8?
For fillet welds longer than 150a (where a = throat thickness), the stress distribution becomes non-uniform due to the flexibility of the connected parts. The effective length is reduced by factor ÃÂòLw = 1.2 - 0.2(Lw/150a) but ÃÂòLw âÃÂÃÂ¥ 0.6. For example, a 900 mm long fillet weld with a = 6 mm: Lw = 900 > 150 ÃÂà6 = 900, so ÃÂòLw = 0.6. This means only 60% of the weld length is considered effective. This is particularly relevant for long stiffener welds in plate girders.
Reference only. Verify all values against the current edition of BS EN 1993-1-8:2005 Clause 4 and UK NA. This information does not constitute professional engineering advice.