A resistance spot-welding operation performed on two pieces of 2.5 mm thick sheet steel uses 12,000 amps for a 0.20 s duration. The electrodes are 6 mm in diameter at the contacting surfaces. Resistance is assumed to be 0.0001 and the resulting weld nugget is 6 mm in diameter and averages 3 mm in thickness. The unit melting energy for the metal Um = 12.0 J/mm3. What portion of the heat generated was used to form the weld nugget, and what portion was dissipated into the work metal, electrodes, and surrounding air?

## EXPERT ANSWER

**Given data:**

Thickness of the work piece is, Z = 2.5 mm

current, I = 12000 A

Time, t = 0.2 s

Diameter of the electrodes, d = 6 mm

Resistance, R = 0.0001 Ohm

diameter of the nugget, dn = 6 mm

thickness of the nugget, Zn = 3 mm

The unit melting energy, q = 12.0 J/mm^{3}

**Solution:**

Volume of the nugget,

Putting the values,

**Vn = 84.82 mm ^{3}**

So, **the total melting energy = unit melting energy * Volume of the nugget**

Q = q * Vn

**Q = 12 * 84.82 = 1017.876 Joules Answer**

**This is the portion of the consumed energy wivh is used to form the weld nugget.**

We know that,

The total consumed heat is,

putting the values,

**H = 2880 Joules**

So, the heat dissipated to the work metal, electrodes and surrounding air is,

Putting the values,

H_{d} = 2880 – 1017.876

**H _{d} = 1862.124 Joules Answer**

**So, the dissipated heat is 1862.124 Joules.**