A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be _____________kN (round off to the nearest integer)

The correct answer is 930.

To determine the roll force required for the rolling operation, we analyze the parameters given in the problem statement and the attached image:
- Plate width, $w=200\text{mm}$
- Initial plate thickness, $h_0=20\text{mm}$
- Final plate thickness, $h_f=18\text{mm}$
- Roll radius, $R=300\text{mm}$
- Strength coefficient, $k=300\text{MPa}$
- Strain hardening exponent, $n=0.2$
- Coefficient of friction, $\mu =0.1$

Step 1: Calculate the true strain (${\epsilon }_T$)
The true strain is determined by the deformation in thickness:
$${\epsilon }_T=\left|\ln \left(\frac{h_f}{h_0}\right)\right|=\left|\ln \left(\frac{18}{20}\right)\right|=\left|-0.10536\right|=0.10536$$

Step 2: Calculate the average flow stress (${\sigma }_0$)
The average flow stress is computed using the material's flow curve equation:
$${\sigma }_0=\frac{k{\epsilon }_T^n}{n+1}$$
Substituting the given values:
$${\sigma }_0=\frac{300\times {\left|\ln \left(\frac{18}{20}\right)\right|}^{0.2}}{1.2}\approx 159.39\text{MPa}$$

Step 3: Calculate the contact length ($L$) and average thickness ($h_{\text{avg}}$)
The projected length of the roll-plate contact arc ($L$) is:
$$L=\sqrt{R\times (h_0-h_f)}=\sqrt{300\times 2}=\sqrt{600}\approx 24.495\text{mm}$$
The average thickness of the plate ($h_{\text{avg}}$) during the pass is:
$$h_{\text{avg}}=\frac{h_0+h_f}{2}=\frac{20+18}{2}=19\text{mm}$$

Step 4: Calculate the roll force ($F$)
Assuming plane strain conditions during rolling, the roll force is given by the formula:
$$F=\frac{2}{\sqrt{3}}\times {\sigma }_0\times L\times w\times \left[1+\frac{\mu L}{4h_{\text{avg}}}\right]$$
Substituting the calculated values into the expression:
$$F=\frac{2}{\sqrt{3}}\times 159.39\times \sqrt{300\times 2}\times 200\times \left[1+\frac{0.1\times \sqrt{300\times 2}}{4\times 19}\right]$$
Evaluating each term step-by-step:
$$F\approx 1.1547\times 159.39\times 24.495\times 200\times \left[1+\frac{2.4495}{76}\right]$$
$$F\approx 901510\times \left[1+0.03223\right]$$
$$F\approx 901510\times 1.03223\approx 930565\text{N}\approx 930.6\text{kN}$$
Rounding to the nearest integer, the roll force is 930 kN.