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Welding Rod Consumption: Calculation Guide

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Key takeaways:

1. The efficiency of welding material usage is a critical factor in project cost management, with the utilization rate of welding wire being higher than that of welding rods, implying that choosing the right welding material can lead to significant savings and reduce waste.

2. Accurate calculations of welding rod consumption hinge upon a variety of factors, including joint type, weld length, and material density, with industry-specific formulas and tables serving as essential tools to prevent inventory excess or shortfall, thereby ensuring project continuity and quality.

3. The provided formulas for electrode consumption demonstrate a nuanced approach to accounting for material loss coefficients and the quality of electrode coatings, reflecting the intricate balance between theoretical calculations and practical adjustments necessary for precise estimation of welding material needs.

To determine the usage of welding rods, a common method used in practice is to calculate the weight of the weld metal, and then divide it by the welding material efficiency.

Calculating the efficiency of welding materials is necessary. However, as the diameter of welding rods and wires differs, the efficiency will also vary greatly.

For the industry, accurately calculating the utilization rate of welding materials can help reduce unnecessary waste in welding.

According to the experience of welders, the utilization rate of welding wire is higher than that of welding rods.

Some industries have a specific recommended table for calculating the weight of welding materials, which is usually divided based on the size of the groove and the number of welding materials required per meter for a certain number of groove degrees (in this case, the utilization rate is typically included).

If such information is available, you can create a spreadsheet, a formula, and then fill it out each time.

It’s crucial to correctly estimate the demand for welding rods during welding construction. If the calculation is too high, there will be overstocked inventory, but if the estimate is too low, the project budget will be insufficient and can even impact the normal progress and quality of the project.

Welding Rod Consumption Formula

The consumption of welding rods is mainly determined by factors such as the joint type, groove type, weld length, and other characteristics of the welding structure.

Here are some calculation formulas:

A) The calculation formula for electrode consumption is as follows:

m = A*L*ρ/1 – KS

Where

  • m — Consumption of welding rod (g);
  • A — Cross-sectional area of weld (cm2); (Refer to Table 1)
  • L — Weld length (cm);
  • ρ — Density of deposited metal (g/cm3);
  • Ks — electrode loss coefficient,

B) The other is the formula for calculating the consumption of non-ferrous powder electrode:

m = ALρ/Kn * (1+Kb)

Where,

  • m — Consumption of welding rod (g);
  • A — Cross-sectional area of weld (cm2); (Refer to Table 1)
  • L — Weld length (cm);
  • ρ — Density of deposited metal (g/cm3):
  • Kb — Coating quality coefficient, see the table below:
  • Kn — Transfer coefficient of metal from electrode to weld (including loss due to burning, splashing and electrode tip)

For example:

1. The steel plate with a known thickness of 20mm is provided with a V-shaped groove, the weld length L is 3m, and the welding rod is 5015. How to calculate the consumption of welding rod?

(According to the table, the sectional area of deposited metal A=250mm2, the density of steel ρ= 7.8g/cm3, conversion coefficient Kn=0.79, weight coefficient of electrode coating Kb=0.32.)

Solution: known sectional area of deposited metal A=250mm2, steel density ρ= 7.8g/cm3, transfer coefficient Kn=0.79, Kb=0.32, L=3m.

From the formula:

mrod = ALρ(1+Kb)/1000Kn = 250mm2×3m×7.8g/cm3(1+0.32)/(1000×0.79)=9.77kg

Answer: The consumption of welding rod is 9.77Kg.

2. Manual arc welding is used to weld a 10m carbon steel fillet weld. The electrode diameter is Φ4.0, and the fillet size is 10mm. How many electrodes are required? (Welding rod deposition rate is 55%)

Requirement of welding material W= D/η=1.2ALρ/η

A=10*10/2=50mm2, L=10m, ρ=7.8*103/kg/m3η=55%

Therefore, W=1.2*(50*10-3)*10*7.8*103/55%=8.509kg≈8.5kg

Answer: 8.5kg Φ4.0 welding rod is required for this weld.

Table 1 Sectional Area of Weld Deposited Metal

NO.Weld nameType and size of welded joint and groove/mmCalculation formula
1Single-side I-shaped weld Sectional Area of Weld Deposited Metal 
2I-shaped weld Sectional Area of Weld Deposited Metal 
3V-shaped weld (no back welding)V-shaped weld (no back welding) Sectional Area of Weld Deposited Metal 
4Single-side V-shaped weld (no back welding)Single-side V-shaped weld (no back welding) Sectional Area of Weld Deposited Metal 
5U-shaped weld (no back welding)U-shaped weld (no back welding) Sectional Area of Weld Deposited Metal 
6Back sealing weld without root overhang at the root of V-shaped and U-shaped weldBack sealing weld without root overhang at the root of V-shaped and U-shaped weld Sectional Area of Weld Deposited Metal 
7Root of V-shaped and U-shaped weldsRoot of V-shaped and U-shaped welds Sectional Area of Weld Deposited Metal 
8Retain V of steel backing plateRetain V of steel backing plate Sectional Area of Weld Deposited Metal 

Table 2 Weight coefficient Kb of electrode coating

E4303E43015E5015
0.770.770.79

Table 3 Transfer coefficient Kn of electrode

E4303E43015E5015
0.42-0.480.42-0.50.38-0.44

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