Arc, gas and resistance welding; heat input

Why it matters

Most steel frames, robot bases, pressure vessels and car bodies are joined by arc welding or resistance spot welding. The heat put into the joint decides penetration, distortion, the width and hardness of the heat-affected zone (HAZ) and the risk of cracking, so engineers specify welding current, voltage and speed through heat input rather than by trial and error.

Key ideas

Fusion welding. The edges of the parts (and usually a filler) are melted and allowed to solidify into one piece. The joint has three regions: the fusion zone (solidified weld metal), the HAZ (base metal heated enough to change its microstructure but not melted), and unaffected base metal.

Arc welding. An electric arc between an electrode and the work reaches about 5000–6000 °C at its core.

• SMAW (manual metal arc, stick) — consumable flux-coated electrode; the coating melts to give shielding gas and slag. Cheap, portable, all positions.
• GMAW (MIG/MAG) — continuous consumable wire fed through the torch with an inert (Ar, He) or active (CO₂, Ar–CO₂) shielding gas; high deposition, easy to automate and to put on a robot.
• GTAW (TIG) — non-consumable tungsten electrode with argon or helium shielding; filler added separately if needed. Clean, precise welds on thin sheet, stainless steel, aluminium and titanium.
• SAW (submerged arc) — wire fed under a blanket of granular flux; very high currents and deposition on thick plate in the flat position.
• Polarity. With DC electrode-positive (DCEP, reverse polarity) about two thirds of the heat is released at the electrode end, which suits consumable electrodes and gives oxide cleaning on aluminium. With DC electrode-negative (DCEN, straight polarity) more heat goes into the work, giving deeper penetration; it is standard for TIG on steel. AC is used for TIG on aluminium.
• Arc characteristics. Arc voltage rises roughly linearly with arc length (V = A + B·l). A drooping (constant-current) power source is used for manual processes, so small changes in arc length barely change the current. A flat (constant-voltage) source with constant wire feed gives a self-regulating arc in GMAW.

Gas welding. Oxy-acetylene gives the hottest flame, about 3100–3200 °C at the tip of the inner cone. A neutral flame (oxygen : acetylene about 1 : 1) suits most steels; an oxidising flame (excess oxygen) is used for brass and bronze; a carburising (reducing) flame (excess acetylene) for high-carbon steel, hard-facing and nickel. Heat input is slow and spread out, so distortion and HAZ width are large; it is now used mainly for repair, thin sheet and brazing. No separate shielding gas is used; the outer flame envelope and flux protect the pool.

Resistance welding. Current is passed through the clamped parts; heat I²Rt is generated mostly at the contact resistance between the sheets, forming a molten nugget under electrode pressure. Variants: spot (overlapping sheets, car bodies), seam (rotating wheel electrodes, leak-tight seams in tanks), projection (current concentrated at embossed projections), flash and upset butt welding (end-to-end joining of bars and rails). Currents are very high (several kA to tens of kA), voltages low (a few volts), and times short (fractions of a second), so no filler, flux or shielding gas is needed.

Heat input and its effects. Heat input is the arc energy delivered per unit length of weld. High heat input gives deeper penetration and slower cooling (less martensite, less hydrogen cracking risk in hardenable steels) but also a wider HAZ, coarser grains, lower toughness and more distortion. Low heat input cools fast, risking hard brittle HAZ and lack of fusion. Arc efficiency η (the fraction of electrical power that reaches the work) depends on the process: roughly 0.75–0.85 for SMAW and GMAW, 0.9 or more for SAW, and about 0.6–0.7 for GTAW; take the exact value from your code or data book.

Formulas

H = η · V · I / v

• H = net heat input per unit length (J/mm), η = arc (heat-transfer) efficiency (–), V = arc voltage (V), I = current (A), v = travel speed (mm/s). With η = 1 this is the gross arc energy.

H (kJ/mm) = 60 · V · I / (1000 · S)

• Same relation with travel speed S in mm/min. Many codes quote heat input this way and then multiply by η.

V = A + B · l

• Linear arc characteristic: l = arc length (mm), A (V) and B (V/mm) are constants for the process and gas.

V = V_o − (V_o / I_s) · I

• Linear (drooping) power-source characteristic: V_o = open-circuit voltage (V), I_s = short-circuit current (A). The operating point is where this line meets the arc characteristic. Arc power is maximum at I = I_s/2, V = V_o/2, giving P_max = V_o · I_s / 4.

Q = I² · R · t

• Resistance-welding heat: Q (J), I (A), R = effective resistance (Ω), t = weld time (s).

η_m = u · A_w · v / (η · V · I)

• Melting efficiency: u = energy to melt unit volume (J/mm³, from a data book), A_w = weld-metal cross-section (mm²), v (mm/s).

Worked examples

Example 1 (standard) — heat input. A GMAW weld uses 24 V, 200 A and travel speed 5 mm/s. Arc efficiency is 0.8. Find gross and net heat input.

1. H_gross = V · I / v = (24 × 200) / 5 = 960 J/mm.
2. H_net = η · H_gross = 0.8 × 960 = 768 J/mm.
3. Check with the mm/min form: 5 mm/s = 300 mm/min; 60 × 24 × 200 / (1000 × 300) = 0.96 kJ/mm ✓.
4. Final: gross 0.96 kJ/mm, net 0.768 kJ/mm.

Example 2 (GATE level) — operating point of the arc. A power source has open-circuit voltage 80 V and short-circuit current 800 A, with a linear characteristic. The arc characteristic is V = 20 + 4l (V, with l in mm). Find current and power for l = 5 mm, then for l = 6 mm.

1. Source: V = 80 − (80/800) I = 80 − 0.1 I.
2. Arc at l = 5 mm: V = 20 + 4 × 5 = 40 V.
3. Equate: 40 = 80 − 0.1 I → I = 400 A. Power P = 40 × 400 = 16 000 W.
4. Arc at l = 6 mm: V = 44 V → I = (80 − 44)/0.1 = 360 A, P = 44 × 360 = 15 840 W.
5. Final: 400 A and 16.0 kW at 5 mm; 360 A and 15.84 kW at 6 mm. At 5 mm the source runs at its maximum-power point (I_s/2, V_o/2), and a 1 mm change in arc length changes the current by 10 %.

Example 3 (resistance spot weld). Current 10 kA flows for 0.2 s through an effective resistance of 100 μΩ. Find the heat generated.

1. Q = I² R t = (10 000)² × 100 × 10⁻⁶ × 0.2.
2. Q = 10⁸ × 10⁻⁴ × 0.2 = 2000 J.
3. Final: Q = 2.0 kJ. Only part of this melts the nugget; the rest is conducted into the sheets and water-cooled electrodes.

Common mistakes

• Mixing travel speed units: mm/s in V·I/v gives J/mm; mm/min needs the factor 60.
• Forgetting the arc efficiency when the question asks for net heat input, or applying it when only arc energy is asked.
• Reversing polarity effects: DCEN puts more heat into the work in TIG; DCEP heats the electrode.
• Saying gas welding uses a shielding gas; it is the flame envelope (and flux) that shields.
• Using I²Rt with milliohms or microohms left unconverted.
• Thinking the arc length does not matter with a drooping source; it changes voltage and slightly changes current.

For GATE ME

Expect heat-input and net-heat numericals, power-source/arc-characteristic intersection problems (operating current, maximum arc power), resistance-welding I²Rt heat with melting energy, and concept questions on polarity, flame types, shielding and matching processes to applications. Practise solving the two straight-line characteristics quickly and keeping units of speed consistent.

Quick check

1. Which flame is used for welding brass, and why?
2. In TIG welding of steel, which polarity gives deeper penetration?
3. Find the heat input for 30 V, 150 A, 250 mm/min (efficiency 1).
4. For V_o = 60 V and I_s = 600 A, what is the maximum arc power?
5. Where is heat concentrated in resistance spot welding?

Answers: 1. Oxidising flame, which forms a zinc-oxide film that limits zinc fuming; 2. DCEN (electrode negative); 3. 60 × 30 × 150 / (1000 × 250) = 1.08 kJ/mm; 4. 60 × 600 / 4 = 9000 W; 5. At the faying (contact) surface between the sheets, where resistance is highest.