Flat rolling draft, roll force and power

Given: $t_0 = 25$ mm, $t_f = 20$ mm, $w = 300$ mm, $R = 300$ mm, $N = 100$ rpm, $\mu = 0.15$, $\bar{Y}_f = 200$ MPa.

Draft actually applied: $d = t_0 - t_f = 5$ mm.

(a) Maximum possible draft

The rolls can bite the strip if $\mu \ge \tan\alpha$, where $\alpha$ is the angle of bite. The maximum draft is

$$d_{max} = \mu^2 R = (0.15)^2\times 300 = 6.75\ \text{mm}$$

The applied draft (5 mm) is less than 6.75 mm, so the pass is possible.

(b) True strain

$$\varepsilon = \ln\frac{t_0}{t_f} = \ln\frac{25}{20} = 0.223$$

(c) Contact length

$$L = \sqrt{R\,(t_0 - t_f)} = \sqrt{300\times 5} = 38.73\ \text{mm}$$

(d) Roll force

The force is the average flow stress times the contact area:

$$F = \bar{Y}_f\, w\, L = 200\times 300\times 38.73 = 2.324\times10^{6}\ \text{N} = 2.32\ \text{MN}$$

(This simple estimate neglects friction. With $\mu = 0.15$ the friction hill would raise the force by roughly 10-15 percent.)

(e) Torque and power

The resultant force acts at the middle of the contact length, so the arm is $L/2$:

$$T = F\,\frac{L}{2} = 2.324\times10^{6}\times 0.01937 = 45.0\ \text{kN\,m per roll}$$

For two rolls the power is

$$P = 2\times T\,\omega = 2\times 45.0\times10^{3}\times \frac{2\pi\times100}{60} = 9.42\times10^{5}\ \text{W}$$

Answer: $d_{max} = 6.75$ mm, $\varepsilon = 0.223$, $L = 38.7$ mm, $F = 2.32$ MN, $T = 45$ kN m per roll, $P = 942$ kW.