Welding cost, filler metal and heat input

It estimates how much wire or electrode to buy, how many welder hours and how much gas a welding job takes, the cost per foot and the heat input of the pass. It starts from the joint geometry (fillet, single-V, double-V or square) to get the weld cross-section and deposited metal, then uses typical deposition efficiency and rate for each process. Use it to quote, buy filler metal and compare processes.

It is the share of the filler metal bought that ends up in the weld. With stick electrodes you lose the stub, the coating turns into slag and there is spatter, so only 55% to 65% of the weight stays in the joint; solid-wire MIG/MAG tops 90%. In the example, 0.58 lb/ft of deposited metal needs 0.61 lb/ft of wire at 95% efficiency.

The example is a 60° single-V groove in 1/2 in plate welded with MIG/MAG: the weld section is 0.17 in², or 0.58 lb of metal per foot. For 150 ft of weld it takes 91.5 lb of wire, 11.3 h of arc time and 25.1 welder hours (at a 45% operating factor), 361 ft³ of gas and 83 kWh. Total cost is $1,709, or $11.40 per foot, and labor is about four fifths of it.

It is the share of the welder’s paid time with the arc on. The rest goes to tacking, cleaning, grinding, changing electrodes and positioning the part. At 45%, the welder welds only 27 minutes an hour, which is why labor dominates weld cost. Measuring the real factor on the shop floor usually corrects an estimate more than any other input.

With the EN 1011-1 / ISO/TR 17671-1 formula: Q = k · U · I · 60 ÷ (1000 · v), with volts, amps and travel speed in mm/min, where k is the process thermal efficiency (0.8 for MIG/MAG, stick and flux-cored, 0.6 for TIG, 1.0 for submerged arc). In the example, 26 V, 240 A and 12 in/min give 25.0 kJ/in. AWS D1.1 uses arc energy without k (31.2 kJ/in here); the allowed range comes from your qualified WPS.

The weld section is geometric: root penetration, back-gouging and back-welding, misalignment and gaps beyond the drawing are not included and usually raise consumption. Efficiency, deposition rate and operating factor are typical starting values. The calculation does not replace procedure or welder qualification under AWS D1.1, nor does it assess preheat.

It depends on weld size and process. For the example joint (60° V in 1/2 in plate, 0.58 lb/ft of deposited metal), stick electrode at about 60% efficiency would need roughly 0.96 lb of electrode per foot, against 0.61 lb/ft of MIG/MAG wire. Pick the process in the list to see the figure with each one’s typical settings.