PEB Studio
Fabrication & welding

Estimating deposited metal and welding consumables

The arithmetic behind kg of wire per tonne of steel: weld cross-section, deposited metal, deposition efficiency and a worked example.

The three-step formula

1. A = cross-section area of the weld (mm²)
2. m = A × L × ρ (deposited metal; ρ ≈ 7,850 kg/m³)
3. consumable = m ÷ η (η = deposition efficiency of the process)

Cross-section area

  • Equal-leg fillet with leg z: A = z² ÷ 2 (a 6 mm leg gives 18 mm²). Add an allowance for convexity as your procedure requires.
  • Butt welds: area of the groove from thickness, bevel angle, root gap and root face, plus the cap reinforcement.

Worked example

1 m of fillet weld with z = 6 mm: A = 18 mm² = 18 × 10⁻⁶ m². m = 18 × 10⁻⁶ × 1 × 7,850 ≈ 0.141 kg of deposited metal per metre. With a solid MAG wire at η ≈ 0.9, about 0.157 kg of wire is needed per metre of weld.

Typical deposition efficiencies

ProcessTypical η (orientation only)
SMAW (stick electrodes, stub losses included)≈ 0.55 – 0.65
FCAW≈ 0.80 – 0.85
GMAW / MAG solid wire≈ 0.90 – 0.95
SAW (wire; flux counted separately)≈ 0.95 – 0.99

These are reference values. Confirm them against the consumable manufacturer's data and your own WPS before using them for orders or prices.

Benchmarks per tonne

Normalising by steel weight makes projects comparable: weld density in metres of weld per tonne (m/t) and consumable rate in kg per tonne (kg/t). In the PEB Studio demo project, 73.18 t of steel carry 6,705 m of welds (91.6 m/t) and need 887.6 kg of consumables (12.13 kg/t).

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