CWI Welding Calculations: Formulas and Worked Examples

What to expect

Part A includes a share of welding-related calculations. The math is basic arithmetic, fractions and a little geometry. The hard part is picking the right formula and not slipping on units. Part A is closed book and 135 minutes long for 165 questions, so these should take you little time. Check the current AWS exam information for what calculator, if any, you may bring, and practice doing these by hand as well.

Work every formula below by hand until you can do it without looking. Wrong answer choices are often the results of common mistakes, such as forgetting a factor or dividing by the wrong number. A careless setup can lead you straight to one of them.

Fractions, decimals and millimeters

Inspectors read rules, gauges and drawings in both fractions and decimals, and in inches and millimeters.

• Fraction to decimal: divide the top by the bottom. 7/16 = 0.4375.
• Inches to millimeters: multiply by 25.4. 1/2 in. = 12.7 mm; 5/8 in. = 15.9 mm.
• Millimeters to inches: divide by 25.4. 6 mm = 0.236 in.

Watch the decimal point. A result ten times too large or too small is a common distractor.

Temperature

• Fahrenheit to Celsius: C = (F - 32) x 5/9. Example: 250 degrees F = (250 - 32) x 5/9 = 121 degrees C.
• Celsius to Fahrenheit: F = C x 9/5 + 32. Example: 200 degrees C = 200 x 1.8 + 32 = 392 degrees F.

The usual mistakes are forgetting to subtract or add 32, adding when you should subtract, and using 5/9 where 9/5 belongs.

Stress units

1 ksi is about 6.895 MPa. A 60 ksi tensile strength is about 414 MPa. If you divide instead of multiplying, you get a number near 9, which can't be a steel's tensile strength in MPa.

Heat input

Heat input (J/in.) = volts x amps x 60 / travel speed (in./min)

The 60 converts minutes to seconds. Example: 30 V, 250 A, 15 in./min gives 30 x 250 x 60 / 15 = 30,000 J/in., or 30 kJ/in.

You may be asked to work backward. Suppose a welding procedure specification (WPS) limits heat input to 45 kJ/in. and the welder runs 26 V and 220 A. The slowest allowed travel speed is 26 x 220 x 60 / 45,000 = 7.6 in./min. Slower travel puts more heat into each inch of weld. Why heat input matters to the steel is covered in the metallurgy and heat control guide.

Travel speed itself is distance over time. A 30 in. weld made in 2 minutes is 15 in./min.

Tension test results

Stress = load / area.

• A round specimen of 0.505 in. diameter has an area of about 0.200 in.^2. A breaking load of 15,000 lbf gives 15,000 / 0.200 = 75,000 psi.
• A reduced-section specimen 1.5 in. wide and 0.5 in. thick has an area of 0.75 in.^2. A failure load of 54,000 lbf gives 54,000 / 0.75 = 72,000 psi.

Common errors are dividing by width or thickness alone, or by the diameter instead of the area. How the tension test itself is run is in the destructive testing guide.

Ductility: elongation and reduction of area

• Percent elongation = (final gauge length - original) / original x 100. A 2.000 in. gauge length that stretches to 2.620 in. gives 0.62 / 2 x 100 = 31%.
• Reduction of area = (original area - final area) / original area x 100. An original area of 0.200 in.^2 that necks to 0.110 in.^2 gives 0.090 / 0.200 x 100 = 45%.

Always divide by the original value.

Fillet weld geometry

For an equal-leg fillet weld with a flat face, the cross section is a right isosceles triangle. The terms leg and throat are defined in the definitions and terminology guide.

• Theoretical throat = leg x 0.707. A 3/8 in. fillet has a throat of about 0.265 in.
• Cross-sectional area = leg x leg / 2. A 5/16 in. fillet has an area of about 0.049 in.^2.

Forgetting to divide by 2 doubles the area, and that doubled value is often among the choices.

Weld metal volume and weight

Weight = cross-sectional area x length x density. Steel weighs about 0.283 lb/in.^3.

Example: a 1/4 in. fillet, 20 ft long.

• Area = 0.25 x 0.25 / 2 = 0.03125 in.^2.
• Length = 240 in. (20 ft x 12).
• Volume = 0.03125 x 240 = 7.5 in.^3.
• Weight = 7.5 x 0.283 = about 2.1 lb.

Convert feet to inches before multiplying. This estimate leaves out reinforcement, spatter and electrode stubs.

Groove area

A single-V groove with no root opening or root face is a triangle. Take plate thickness t and groove angle A. The top width is 2 x t x tan(A/2). The area is t x t x tan(A/2).

Take 3/4 in. plate and a 60-degree groove. The area is 0.75 x 0.75 x 0.577 = 0.325 in.^2. Root openings and reinforcement add more.

Carbon equivalent

One common formula:

CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15

Example: 0.18% C, 1.20% Mn, 0.20% Cr, 0.10% Mo, 0% V, 0.20% Ni, 0.10% Cu.

• C = 0.18.
• Mn/6 = 0.20.
• (Cr + Mo + V)/5 = 0.30 / 5 = 0.06.
• (Ni + Cu)/15 = 0.30 / 15 = 0.02.
• CE = 0.46.

Keep each term separate and add at the end. Leaving out a term is an easy error to make.

Percentages from acceptance criteria

Codes often state limits as percentages. Say an undersized fillet is allowed for up to 10% of the weld length. Then a 60 in. weld may have 6 in. of undersize. Read the code to see whether the limit applies per weld or per specified length.

A routine for every problem

1. Write the formula before plugging in numbers.
2. Convert units first: feet to inches, minutes to seconds.
3. Estimate the size of the answer before calculating.
4. Compare your result with the choices, and ask which mistake produced each wrong one.

Mark J. Curry's tip: is your answer exactly double, half or ten times one of the choices? Then check the step most likely to cause that factor.

Work through the free welding calculation practice questions. To see which Part A topic to study first, take the free CWI diagnostic test.

Sources

The Welding-Related Calculations questions cite these documents, by document and, where we can confirm it, chapter or section. We never reproduce them.

1. AWS D1.1 (2 questions)
2. AWS A3.0 (1 question)

Found a mistake? Use "Report a problem" under any question, and Mark J. Curry checks it.

Questions? support@cwiready.com