Question

The thickness of a steel plate with material strength coefficient of 210 MPa, has to be reduced from 20 mm to 15 mm in a single pass in a two-high rolling mill with a roll radius of 450 mm and rolling velocity of 28 m/min. If the plate has a width of 200mm and its strain hardening exponent. n is 0.25, the rolling force required for the operation is kN (round off to 2 decimal places)

Note: Average Flow Stress = Material Strength

$Coefficient\times \frac{(Truestrain{)}^n}{(1+n)}$

Answer 1

True strain, $\left({ϵ}_T\right)=In\left(\frac{h_f}{h_0}\right)=In\left(\frac{15}{20}\right)=-0.2876$

$\overline{{\sigma }_0}=K\left(\frac{{ϵ}_T^n}{1+n}\right)=210\times \frac{(0.28768{)}^{0.25}}{1+0.25}=123.04MPa$

F = $\overline{{\sigma }_0}\times \sqrt{R\Delta h}\times b$

= $123.04\times \sqrt{450\times (20-15)}\times 200$

= 1167259.9 N = 1167.26 kN