Welding math
Fillet Weld Area, Volume, and Weight
Estimating filler metal starts with the area of a triangle: half the leg squared gives the cross-section, length turns it into volume, and density turns volume into the pounds a job will actually buy.
The formula
A = leg^2 / 2, Vol = A x L, W = Vol x density
The cross-sectional area of an ideal equal-leg fillet is the leg squared divided by two; multiply by weld length for volume, and by material density for deposited weight.
What each symbol means
| Symbol | Meaning | Units |
|---|---|---|
| A | Cross-sectional area of the ideal flat-faced fillet | in^2 |
| leg | Leg length of the equal-leg fillet | in |
| L | Total length of weld to be deposited | in |
| W | Weight of deposited weld metal | lb |
| rho | Density of the deposit, 0.284 for carbon and low alloy steel | lb/in^3 |
Why it works
An ideal equal-leg fillet is a right triangle with both legs the same, so its area is half the base times the height, which collapses to leg squared over two. That single expression is behind almost every filler metal estimate on structural work. Note how sharply it scales: doubling the leg quadruples the area, so a half-inch fillet consumes four times the metal of a quarter-inch fillet over the same run.
Volume follows immediately by multiplying the cross-section by the total length of weld, and weight follows by multiplying volume by density. Carbon and low alloy steel run about 0.284 pounds per cubic inch. Stainless is slightly denser, aluminium is roughly a third the weight, so an aluminium estimate built on a steel density will be wrong by a factor of three.
The result is a floor, not a forecast. It counts only the ideal triangle, so anything the welder adds on top of it is extra. Real beads carry convexity, welders overwelding by even a sixteenth of an inch of leg add a measurable fraction to the total, and none of the metal lost to slag, spatter, or stubs appears here. Estimating purchases means starting with this number and then dividing by a deposition efficiency.
Worked examples
Filler metal for a run of quarter-inch fillet
- Leg size
- 0.25 in
- Total weld length
- 120 in
- Steel density
- 0.284 lb/in^3
- Cross-sectional area: 0.25 x 0.25 / 2 = 0.03125 in^2.
- Volume over the run: 0.03125 x 120 = 3.75 in^3.
- Weight of deposit: 3.75 x 0.284.
Result: 0.03125 in^2 of section, 3.75 in^3 of metal, 1.065 lb deposited
Ten feet of quarter-inch fillet is barely a pound of weld metal in the joint. Once deposition efficiency is applied, the electrode actually consumed will be noticeably more than that.
Doubling the leg on a longer run
- Leg size
- 0.5 in
- Total weld length
- 240 in
- Steel density
- 0.284 lb/in^3
- Cross-sectional area: 0.5 x 0.5 / 2 = 0.125 in^2, four times the quarter-inch section.
- Volume: 0.125 x 240 = 30 in^3.
- Weight: 30 x 0.284.
Result: 0.125 in^2 of section, 30 in^3 of metal, 8.52 lb deposited
Twice the leg and twice the length is eight times the metal, not four. This is the arithmetic behind the standing advice never to overweld a fillet, because the cost penalty compounds.
In practice
- Add a convexity allowance before ordering. A flat-face triangle underestimates a real bead, and shops commonly carry a modest percentage on top of the ideal figure for that reason.
- Overwelding is invisible on a drawing and expensive on an invoice. A sixteenth of extra leg on a quarter-inch fillet is a quarter more metal, plus the arc time to put it there.
- Deposited weight is not purchased weight. Divide by deposition efficiency to get the electrode a job actually consumes.
- For a concave fillet the ideal triangle overestimates the metal present, which is a warning about strength rather than a saving on the estimate.
Related topics
Practise the arithmetic under pressure
The welding-math domain in the question bank is where these formulas get tested the way they get tested on paper — with distractors built from the mistakes people actually make.