Question

In a cold rolling process without front and back tensions, the required minimum coefficient of friction is 0.04. Assume large rolls. If the draft is doubled and roll diameters are halved, then the required minimum coefficient of friction is ___________. (Rounded off to two decimal places) 
 

Hint:

For rolling problems, remember the relationship \(\mu_{min} = \sqrt{\Delta h / R}\). This allows you to quickly analyze how changes in draft (\(\Delta h\)) or roll radius (\(R\)) affect the required friction.

Answer 1

Step 1: Identifying the Governing Physics:
In rolling, the condition for the rollers to spontaneously "grab" the metal strip is that the friction angle must be greater than or equal to the bite angle. The limiting condition for the minimum coefficient of friction ($\mu$) is expressed by: \[ \mu = \tan(\alpha) \approx \sqrt{\frac{\Delta h}{R}} \] where $\Delta h$ is the draft (reduction in thickness) and $R$ is the radius of the rolls.
Step 2: Setting up the Proportionality:
From the formula, we see that $\mu \propto \sqrt{\Delta h}$ and $\mu \propto \frac{1}{\sqrt{R}}$. Therefore, the relationship between initial and final states is: \[ \frac{\mu_2}{\mu_1} = \sqrt{\frac{\Delta h_2}{\Delta h_1} \times \frac{R_1}{R_2}} \]
Step 3: Substituting the Given Changes:
The problem states the draft is doubled ($\Delta h_2 = 2\Delta h_1$) and the roll diameter is halved. Since Diameter ($D$) is $2R$, halving the diameter also means halving the radius ($R_2 = 0.5R_1$). \[ \frac{\mu_2}{0.04} = \sqrt{\frac{2\Delta h_1}{\Delta h_1} \times \frac{R_1}{0.5R_1}} = \sqrt{2 \times 2} = \sqrt{4} = 2 \]
Step 4: Final Result:
\[ \mu_2 = 2 \times 0.04 = 0.08 \]