Question

Spot welding of two steel sheets each 2 mm thick is carried out successfully by passing 4 kA of current for 0.2 seconds through the electrodes. The resulting weld nugget formed between the sheets is 5 mm in diameter. Assuming cylindrical shape for the nugget, the thickness of the nugget is _______ mm.

Latent heat of fusion for steel	1400 kJ/kg
Effective resistance of the weld joint	200 μΩ
Density of steel	8000 kg/m3

Answer 1

Concept:

Calculate the heat supplied for melting and heat required for melting.

And then equating both of them will give you the thickness of nugget.

Calculation:

Given:

thickness of steel sheets = 2 mm.

I = 4 kA, time (t) = 0.2 seconds.

Diameter of Nugget = 5 mm

Assuming cylindrical shape of Nugget.

Volume \(= \frac{\pi }{4}{D^2}{t_n}\), where t_n → Thickness of Nugget

Now,

Heat required to melt the above volume of Nugget = Heat required to melt 1 kg × Total mass (m)

∵ density (ρ) = mass/Volume

∴ m = ρ × volume

∴ Heat required to melt volume of nugget (Q)= Heat required to melt 1 kg × ρ × volume

\(\therefore {\rm{Q}} = \left( {1400} \right) \times \frac{{\rm{\pi }}}{4}{\rm{\;}}{{\rm{D}}^2}{{\rm{t}}_{\rm{n}}} \times \left( {8000} \right){\rm{\;kJ}}\)

And

Heat supplied = I²Rt= (4× 10³)² × (200 × 10^-6) × 0.2

∴ Heat supplied = 640 J.

Equating the heat calculated by two methods:

\(640 = \left( {1400} \right)\frac{\pi }{4}{\left( {0.005} \right)^2}{t_n}\left( {8000} \right) \times {10^3}\)

640 = (219.911 × 10³) t_n

t_n = 2.91 × 10^-3 m

∴ t_n = 2.91 mm.