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Question

Titanium strips of 2 mm thickness are joined by resistance spot welding process by applying an electric current 'I' for 0.2 sec. The heat required for melting titanium is \( 5.2~ \text{J/mm}^3\) and the diameter and thickness of weld nugget are found to be 8 mm and 3 mm respectively. Assuming electrical resistance to be \(50~ \mu\Omega\) and total heat generated is used for welding operation, determine the current "I".

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Solution

\(H_m=u\times\text{volume of weld nugget}\)

\(=5.2\times\dfrac{\pi}{4}\times8^2\times3\)

\(H_m=784.14~J\)

\(Also ~~H_m=H_s~~[given]\)

\(\therefore784.14=I^2\times50\times10^{-6}\times0.2\)

\(\Rightarrow~~~~~~I=8855~A\)

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