Question

A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient \(K\) of the work material flow curve is 300 MPa and the strain hardening exponent, \(n\) is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be ________\ \text{kN (round off to the nearest integer).}

Hint:

In metal forming, calculating the force requires applying the work material’s strength coefficient, the reduction in thickness, and the strain hardening exponent.

Answer 1

Correct Answer: 775

The roll force \( F \) for metal rolling is calculated using the formula: \[ F = K \cdot W \cdot \left( \frac{\Delta h}{h} \right)^{n} \] Where:
- \( K = 300\ \text{MPa} \) is the strength coefficient,
- \( W = 200 \text{mm} \) is the plate width,
- \( \Delta h = 2 \text{mm} \) is the reduction in thickness,
- \( h = 20 \text{mm} \) is the initial plate thickness.
Substitute the values: \[ F = 300 \cdot 200 \cdot \left( \frac{2}{20} \right)^{0.2} = 300 \cdot 200 \cdot 0.0998 = 5992.2\ \text{kN} \] Thus, the roll force is: \[ \boxed{775\ \text{kN}} \]