Question

A 80 mm thick steel plate with 400 mm width is rolled to 40 mm thickness in 4 passes with equal reduction in each pass, by using rolls of 800 mm diameter. Assuming the plane-strain deformation, what is the minimum coefficient of friction required for unaided rolling to be possible?

1. 0.111
2. 0.158
3. 0.223
4. 0.316

Answer 1

Correct Answer - Option 2 : 0.158

For maximum reduction formula is:

(DH)_max = μ².R

Where, μ = coefficient of frication

R = Roller radius

Initial thickness = 80 mm, final thickness = 40 mm

Total reduction = 80 - 40 = 40 mm

\(\begin{array}{l} {\rm{Reduction\;per\;pass}} = \frac{{40}}{4} = 10{\rm{\;mm\;}}\left( {{\rm{no}}.{\rm{\;of\;passes}} = 4} \right)\\ \Rightarrow {\rm{DH}} = 10{\rm{\;mm\;}}\& {\rm{\;R}} = \frac{{800}}{2} = 400\;mm \end{array}\)

Plug-in those two values is equation

\({\rm{\mu }} = \sqrt {\frac{{10}}{{400}}} \Rightarrow {\rm{\mu }} = 0.1581\)