A 50mm thick strip rolled to 30mm thickness in 3 passes with equal reduction in each pass by using rolls of 600mm diameter, what is the minimum coefficient of friction required for rolling operation________.Correct answer is between '0.13,0.17'. Can you explain this answer?

Question

Answer

Solution:

Given,
Initial thickness (t1) = 50 mm
Final thickness (t2) = 30 mm
Reduction in thickness in each pass = (t1 - t2)/3 = (50 - 30)/3 = 6.67 mm
Diameter of rolls (D) = 600 mm

We know that the minimum coefficient of friction required for rolling operation can be calculated as:

μmin = (t1 - t2)/(t1 - t2 + 2h)

Where h is the height of the contact zone between the rolls and the strip.

To calculate h, we can use the following formula:

h = (D/2) x √[(t1 - t2)/2]

Therefore, h = (600/2) x √[(50 - 30)/2] = 173.2 mm

Substituting the values of t1, t2, and h in the formula of μmin, we get:

μmin = (50 - 30)/(50 - 30 + 2 x 173.2) = 0.15

Therefore, the minimum coefficient of friction required for rolling operation is 0.15, which lies between the given range of 0.13 and 0.17.

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