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 200 mm 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 : $\mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{F}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}=\mathrm{M}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{C}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{e}\mathrm{n}\mathrm{t}\times \frac{{\left(\mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}\right)}^{\mathrm{n}}}{\left(1+\mathrm{n}\right)}$

The correct answer is 1167.26.

To find the rolling force required for the operation, we follow a step-by-step calculation using the principles of metal rolling:

1. Identify the given parameters from the problem:
Initial thickness, $h_o=20\text{mm}$
Final thickness, $h_f=15\text{mm}$
Roll radius, $R=450\text{mm}$
Plate width, $w=200\text{mm}$
Material strength coefficient, $K=210\text{MPa}$
Strain hardening exponent, $n=0.25$

2. Calculate the True Strain (${ϵ}_T$):
The true strain (magnitude) in the thickness direction during the rolling pass is given by:

$${ϵ}_T=\ln (\frac{h_o}{h_f}=\ln (\frac{20}{15}\approx 0.28768$$

3. Calculate the Average Flow Stress (${\overline{\sigma }}_o$):
Using the formula provided in the question:

$${\overline{\sigma }}_o=K\times \frac{{{ϵ}_T}^n}{1+n}$$

Substituting the values:

$${\overline{\sigma }}_o=210\times \frac{{\left(0.28768\right)}^{0.25}}{1+0.25}=210\times \frac{0.73238}{1.25}\approx 123.04\text{MPa}$$

4. Calculate the Contact Length ($L$):
The contact length between the roll and the plate is calculated using the roll radius ($R$) and the draft ($\mathrm{\Delta h}=h_o-h_f$):

$$L=\sqrt{R\times (h_o-h_f)}=\sqrt{450\times (20-15)}=\sqrt{2250}\approx 47.434\text{mm}$$

5. Calculate the Rolling Force ($F$):
The roll force is the product of the average flow stress, the contact area (width multiplied by contact length):

$$F={\overline{\sigma }}_o\times L\times w$$

Substituting the values:

$$F=123.04\text{N/mm}2\times 47.434\text{mm}\times 200\text{mm}$$

$$F\approx 1167259.9\text{N}=1167.26\text{kN}$$