GATE 2020 ME (ME1) – Question 56
A steel plate with strength coefficient 210 MPa is reduced from 20 to 15 mm in one pass through a two-high rolling mill. Roll radius is 450 mm, rolling velocity 28 m/min, plate width 200 mm and strain-hardening exponent n = 0.25. Find rolling force (kN), to two decimals. Use average flow stress = strength coefficient × (true strain)$^n$/(1+n).
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 1164 to 1168
Explanation
True strain is $\epsilon=\ln(20/15)=0.287682$. Mean flow stress is $\bar\sigma=210\epsilon^{0.25}/1.25$ MPa and contact length is $l=\sqrt{450(20-15)}$ mm. Rolling force is $F=\bar\sigma(200)l/1000=\mathbf{1167.24}$ kN. The simplified model neglects additional friction corrections; rolling speed is not needed.