Calculations for Welding Pipe

Assume the rolled 72"sections are 2 mtr in length

Welding volume calculation Weld V60 º

Plate thickness 10 mm

Root opening 3 mm

Nose 3 mm

Welding volume for V welds 60 ° for each mtr

Rootopening x Plate thickness + ( Plate th. – nose length) ^2 / (SQRT(3) = mm2 cross section

mm2 crossection x 1000 mm length / 10^3 = netto weldvolume cm3

3 x 10 + 49/1,732 = 28,29 mm x 1000 mm / 10^3 = 58,29 cm3

100 pcs of longitudinal weld of 2 mtr 11.658 cm3

99 pcs of circumferential welds 72" 33.083 cm3

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Total nett welding volume 44.741 cm3

Add 10% for spherical closing 4.474 cm3

Add 5% for grind back and counterweld the root 2.237 cm3

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51.452 cm3

= approx 400 kg netto welding metal

Hi Peter,

Thanks a lot for your reply with worked out example.I have been searching for this stuff for the last 6 months.I have raised this query in many forums and groups and unfortunately none of them responded properly.Many people gave answers in a vauge manner as they are the new generation Einsteen!!!!Like every engineer, I also know the common mathematical principle what we have to apply over here.But a worked out example is a different one with unit conversion and it really giving practical exposure.You have done a great job for fresh or less experienced welding engineers.

Again thank you

Regrds,

Inconel

Consider the weld joint design is "V" {Buttweld with 1.6mm gap}. Consider this to be a "Triangle". You need to calculate the area of this Triangle. i.e. the plate you have considered is 10mm thk. Hence, the triangle dimensions - height =10mm, Bevel angle = 30 or 45 deg.. Now calculate the base. Now you have the dimensions of the triangle. Calculate the area of the triangle. Multiply this with total length. This area is what you should be filling with an electrode. {A}

The electrode is of 250 or 350 or 450 mm in length depending upon the electrode dia. Consider 75% of metal deposit efficiency {exclude the wastages}. This would give you weld metal deposit with one electrode. {B}

Now do the simple arithmetic - how much electrodes are required for the total worked out for a given weld design

trust this clarifies.

Hi Sre eram,

If possible provide a worked out example instead of explaing common maths.Please avoid explaining common mathematical principle as we are well aware of those things.We are looking for real practical explanation with examples.