1. Heat input: formula and the efficiency factor convention
Electrical power in the arc is volts × amps, in watts (joules per second). Divide by travel speed to get energy per unit of weld length. With travel speed in inches per minute, multiply by 60 to turn seconds into minutes:
- Arc energy (J/in) = V × A × 60 ÷ travel speed (in/min). Divide by 1,000 for kJ/in.
- Metric: arc energy (kJ/mm) = V × A × 60 ÷ (travel speed (mm/min) × 1,000), as printed by the Australian Welding Institute.
- Heat input = k × arc energy, where k is the thermal efficiency factor. TWI and the Australian Welding Institute both print k = 1.0 for SAW, 0.8 for SMAW, GMAW and FCAW, and 0.6 for GTAW.
Which one you use depends on the document. TWI notes that a number of American codes use the term heat input for what European practice calls arc energy, that is, with no efficiency factor. ASME Section IX, QW-409.1, the heat input variable for procedure qualification, is written that way. If your code, WPS or exam question gives no efficiency factor, use k = 1. Apply k only when the governing document calls for it.
Worked example: 25 V, 200 A, 10 in/min. Power = 5,000 W. Arc energy = 5,000 × 60 ÷ 10 = 30,000 J/in = 30 kJ/in (1.1811 kJ/mm). With k = 0.8 for SMAW, heat input = 24 kJ/in (0.9449 kJ/mm). In metric, 25 V, 200 A at 250 mm/min gives 1.2 kJ/mm; at 4 mm/s it gives 5,000 ÷ 4 = 1,250 J/mm = 1.25 kJ/mm, with no 60 because the speed is already per second.
2. IIW carbon equivalent
CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15, every element in weight percent. The International Institute of Welding adopted it as a simplified hardenability formula; see TWI and the Australian Welding Institute. AWS D1.1 uses a modified form with a silicon term in an annex (the annex letter varies by edition), so check which formula your code names.
Worked example: C 0.18, Mn 1.20, Cr 0.10, Mo 0.02, V 0.01, Ni 0.10, Cu 0.15. CE = 0.18 + 0.200 + 0.13/5 (0.026) + 0.25/15 (0.0167) = 0.423. Add Cr, Mo and V before dividing by 5. What a given CE means for preheat is set by the governing code, not by this calculator.
3. Tensile stress from load and area
Stress = load ÷ original cross-sectional area. A round bar's area is π × d² ÷ 4; a strap's is width × thickness. In US units lbf ÷ in² gives psi; in SI, N ÷ mm² gives N/mm², which is the same as MPa. ASTM E8/E8M is the usual test method for metals; use your own copy for specimen dimensions.
Worked example: a 0.505 in diameter bar breaks at 12,000 lbf. Area = π × 0.505² ÷ 4 = 0.200296 in². Stress = 12,000 ÷ 0.200296 = 59,911 psi (59.9 ksi, 413.07 MPa). In SI, a 12.5 mm bar at 50 kN: area 122.7185 mm², stress 50,000 ÷ 122.7185 = 407.4 MPa.
4. Percent elongation
Elongation (%) = (final gauge length − original gauge length) ÷ original gauge length × 100. Measure the final length between the same gauge marks after fitting the broken halves together.
Worked example: 2.000 in gauge length, 2.460 in after fracture. (2.460 − 2.000) ÷ 2.000 × 100 = 23.0%. Dividing by the final length gives 18.7%, a common wrong answer.
5. Percent reduction of area
Reduction of area (%) = (original area − final area) ÷ original area × 100. For a round specimen this is (1 − (df ÷ d0)²) × 100, measured at the smallest diameter of the fracture.
Worked example: 0.505 in original diameter, 0.300 in at fracture. (0.300 ÷ 0.505)² = 0.3529, so reduction of area = 64.71%. Subtracting the diameters directly gives 40.59%, which is wrong.
6. Inch/mm and psi/MPa conversion
1 in = 25.4 mm exactly. 1 psi is 1 lbf per square inch: 4.4482216 N ÷ 645.16 mm² = 0.0068948 MPa, so 1 MPa = 145.04 psi and 1 ksi = 6.8948 MPa. These match NIST SP 811, Appendix B.8.
Worked examples: 3/8 in = 0.375 × 25.4 = 9.525 mm. 70 ksi = 70 × 6.8948 = 482.63 MPa. 250 MPa = 250 × 145.04 = 36,259 psi.
Common units traps
| Calculation | Trap | Check |
|---|---|---|
| Heat input | Leaving out × 60 with speed in in/min, or reporting J/in as kJ/in | Typical arc welding values are tens of kJ/in, not thousands |
| Heat input | Applying an efficiency factor the code does not use | Use k = 1 unless your document calls for k |
| Carbon equivalent | Entering 0.0018 for 0.18% carbon, or dividing only Cr by 5 | Weight percent; sum Cr + Mo + V first |
| Tensile stress | Radius in place of diameter in π d² ÷ 4, or using the necked area | Original diameter, original area |
| Tensile stress | Mixing psi and ksi | 1 ksi = 1,000 psi; 1 N/mm² = 1 MPa |
| Elongation | Dividing by the final length | Always divide by the original gauge length |
| Reduction of area | Subtracting diameters instead of areas | Square the diameter ratio |
| Conversion | Multiplying MPa by 6.895 to get psi | MPa × 145.04 = psi; ksi × 6.895 = MPa |
These examples are original arithmetic, not exam questions. The worked methods follow the published formulas cited above; the governing code, manufacturer and procedure control every real decision.
Ready to see where you stand across all of Part A? Start from the CWI Part A practice page.