Question

A 50 mm flat aluminium plate is reduced in thickness to 25 mm in a single pass cold-rolling operation with the rolling diameter of 1250 mm. For this case, the minimum required coefficient of friction between the plate and the roll (rounded off to two decimal places) is _______.

Hint:

Use the rolling bite condition \(\mu_{min} = \sqrt{\Delta h / R}\), with \(\Delta h\) the draft (thickness reduction) and \(R\) the roll radius.

Answer 1

Correct Answer: 0.19

Step 1: Recall the bite condition for cold rolling.
For the rolls to grip a plate and pull it through the gap without the plate slipping back out, the friction force at the contact must be at least strong enough to overcome the normal force pushing the plate away from the rolls. Working through this force balance at the point where the roll first touches the plate gives the "angle of bite" condition,
\[ \mu \geq \tan\alpha \]
where \(\mu\) is the coefficient of friction between the plate and the roll, and \(\alpha\) is the bite angle, the angle at the roll center between the entry point of the plate and the point directly below the roll center.

Step 2: Relate the bite angle to the draft and the roll radius.
Let \(R\) be the roll radius and \(\Delta h = h_0 - h_f\) be the draft, the total reduction in plate thickness. From the roll-bite geometry,
\[ \cos\alpha = \frac{R - \Delta h/2}{R} = 1 - \frac{\Delta h}{2R} \]
For the small bite angles seen in practice, \(\cos\alpha \approx 1 - \dfrac{\alpha^2}{2}\). Equating the two expressions for \(\cos\alpha\),
\[ \frac{\alpha^2}{2} \approx \frac{\Delta h}{2R} \quad\Rightarrow\quad \alpha \approx \sqrt{\frac{\Delta h}{R}} \]
Since \(\alpha\) is small, \(\tan\alpha \approx \alpha\), so the minimum coefficient of friction needed to bite the plate is
\[ \mu_{min} = \sqrt{\frac{\Delta h}{R}} \]

Step 3: Read off the given numbers.
The plate goes from \(h_0 = 50\) mm to \(h_f = 25\) mm, so the draft is
\[ \Delta h = 50 - 25 = 25 \text{ mm} \]
The roll diameter is \(1250\) mm, so the roll radius is
\[ R = \frac{1250}{2} = 625 \text{ mm} \]

Step 4: Substitute and compute.
\[ \mu_{min} = \sqrt{\frac{25}{625}} = \sqrt{0.04} = 0.2 \]

Step 5: Final Answer.
Rounded off to two decimal places, the minimum coefficient of friction required is \(0.20\), which lies inside the accepted range of 0.19 to 0.21.
\[ \boxed{\mu_{min} = 0.20} \]