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: Understanding the Question:
The question describes a cold rolling process and asks how the required minimum coefficient of friction changes when the draft and roll diameter are altered.
Step 2: Key Formula or Approach:
In a rolling process, for the material to be drawn into the rolls, the bite angle (\(\alpha\)) must be less than or equal to the friction angle (\(\lambda\)). The minimum required coefficient of friction (\(\mu\)) corresponds to the limiting condition where \(\mu = \tan(\alpha)\). For small angles, \(\tan(\alpha) \approx \alpha\). The bite angle is related to the draft (\(\Delta h\)) and the roll radius (\(R\)) by the formula: \[ \alpha \approx \sqrt{\frac{\Delta h}{R}} \] Therefore, the required minimum coefficient of friction is given by: \[ \mu = \sqrt{\frac{\Delta h}{R}} \] Step 3: Detailed Explanation:
Let the initial conditions be denoted by subscript 1 and the final conditions by subscript 2. 1. Initial Condition: \[ \mu_1 = 0.04 \] \[ \mu_1 = \sqrt{\frac{\Delta h_1}{R_1}} \] So, \(0.04 = \sqrt{\frac{\Delta h_1}{R_1}}\). 2. Final Condition: The problem states:

• The draft is doubled: \(\Delta h_2 = 2 \Delta h_1\)
• The roll diameters are halved, which means the roll radii are also halved: \(R_2 = \frac{R_1}{2}\)

Now, we calculate the new required minimum coefficient of friction, \(\mu_2\): \[ \mu_2 = \sqrt{\frac{\Delta h_2}{R_2}} \] Substitute the new values in terms of the initial ones: \[ \mu_2 = \sqrt{\frac{2 \Delta h_1}{R_1/2}} = \sqrt{4 \times \frac{\Delta h_1}{R_1}} \] \[ \mu_2 = 2 \times \sqrt{\frac{\Delta h_1}{R_1}} \] Since \(\mu_1 = \sqrt{\frac{\Delta h_1}{R_1}}\), we can write: \[ \mu_2 = 2 \times \mu_1 \] \[ \mu_2 = 2 \times 0.04 = 0.08 \] Step 4: Final Answer:
The required minimum coefficient of friction is 0.08.