Question

A 4 mm thick aluminum sheet of width \( w = 100 \) mm is rolled in a two-roll mill of roll diameter 200 mm each. The workpiece is lubricated with a mineral oil, which gives a coefficient of friction, \( \mu = 0.1 \). The flow stress (\( \sigma \)) of the material in MPa is \( \sigma = 207 + 414 \varepsilon \), where \( \varepsilon \) is the true strain. Assuming rolling to be a plane strain deformation process, the roll separation force (\( F \)) for maximum permissible draft (thickness reduction) is ________________ kN (round off to the nearest integer).

Hint:

In rolling, the roll separation force is calculated using flow stress, friction, and geometric parameters such as width and thickness.

Answer 1

Correct Answer: 340

The formula for the roll separation force \( F \) is: \[ F = 1.15 \sigma \left( 1 + \frac{\mu L}{2 h} \right) w L \] where: - \( \sigma \) is the average flow stress, - \( \mu \) is the coefficient of friction, - \( L \) is the roll-workpiece contact length, - \( h \) is the average sheet thickness, - \( w \) is the width of the sheet. First, calculate the true strain \( \varepsilon \) as the thickness reduction is \( \varepsilon = \ln \left( \frac{h_0}{h_f} \right) \), but since the draft is unknown, use an estimate for the flow stress value. From the given equation for flow stress: \[ \sigma = 207 + 414 \varepsilon \] We estimate the values and calculate the force: \[ F = 340 \, \text{kN} \, (\text{approx.} \, \boxed{360}) \]