Weld Bead Volume & Filler Metal Weight

Estimate filler metal consumption in kg or lb for any weld joint cross-section

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Knowing how much filler metal a job will consume drives both the purchase order and the labour estimate. This tool computes the cross-sectional area of the most common joint shapes, scales it by weld length and filler density to get deposited weight, then adds back process losses to give the weight you actually need to buy.

How it works

Every estimate starts from the bead cross-section. The areas, with all dimensions in millimetres, are:

Fillet (equal leg z):   A = 0.5 x z^2 x 1.10      (10% convexity)
V-groove:               A = root x t + t^2 x tan(angle/2)
U-groove:               A = root x t + R x t
J-groove:               A = root x t + 0.5 x R x t

where t is groove depth, R is the root radius, and angle is the included groove angle. Multiplying area by weld length gives the deposited volume, and volume times density gives deposited weight:

weight = area x length x density
purchased = deposited / deposition_efficiency

Example and notes

An 8 mm fillet, 1000 mm long, in carbon steel: area is 0.5 x 8^2 x 1.10 = 35.2 mm^2, volume is 35.2 cm^3, deposited weight is about 0.276 kg. With MIG at 90 percent efficiency you would buy roughly 0.307 kg of wire.

  • The 10 percent fillet allowance reflects a normal convex face; flat or concave beads consume less.
  • Densities used: carbon steel 7.85, stainless 8.00, aluminium 2.70 grams per cubic centimetre.
  • For multi-pass grooves the total area is what matters, so enter the full groove depth rather than a single pass.
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