---
title: "Flat rolling draft, roll force and power"
subject: "Manufacturing and Production Processes"
chapter: 5
key: "semester_4/manufacturing_and_production_processes:flat-rolling-force-power-numerical"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/flat-rolling-force-power-numerical
site: BCT Engineering
---

# Flat rolling draft, roll force and power

Manufacturing and Production Processes · Chapter 5: Bulk Deformation Process

- Asked in: year not given, 8 marks
- Video: [Mechanics of Rolling Operation Part II](https://www.youtube.com/watch?v=EUx1SNzArq8) — Sunil Mohod
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/flat-rolling-force-power-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-5

## Question

A 300 mm wide strip of 25 mm thickness is hot rolled to 20 mm thickness in a single pass by rolls of 600 mm diameter rotating at 100 rpm. The coefficient of friction is 0.15 and the average flow stress of the metal is 200 MPa. Determine (a) the maximum possible draft, (b) the true strain, (c) the roll-strip contact length, (d) the roll force, (e) the torque per roll and the total power.

## Short answer

Maximum draft = μ²R = 6.75 mm, true strain = ln(25/20) = 0.223, contact length = √(300×5) = 38.7 mm. Roll force = 200×300×38.73 = 2.32 MN, torque per roll = 2.32×0.01937 = 45.0 kN m, total power = 2×45×10³×π×100/30 = 942 kW.

## Answer

**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

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

| Quantity | Result |
|---|---|
| Maximum draft | 6.75 mm |
| True strain | 0.223 |
| Contact length | 38.7 mm |
| Roll force | 2.32 MN |
| Torque per roll | 45.0 kN m |
| Total power | 942 kW |

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Maximum possible draft is d_max =
   A. mu R
   B. mu^2 R
   C. R/mu
   D. mu sqrt(R)
   Answer: B. From the bite condition mu >= tan(alpha): d_max = mu^2 R = 0.0225 x 300 = 6.75 mm. (Key point: Maximum draft)

2. Roll-strip contact length for R = 300 mm and draft 5 mm is
   A. sqrt(300 x 5) = 38.7 mm
   B. 300 x 5 = 1500 mm
   C. 5/300 mm
   D. sqrt(600 x 25) mm
   Answer: A. L = sqrt(R (t0 - tf)) = sqrt(1500) = 38.73 mm. (Key point: Contact length)

3. Roll force is computed as
   A. Flow stress x strip thickness
   B. Average flow stress x width x contact length
   C. Flow stress x roll diameter
   D. True strain x width
   Answer: B. F = 200 x 300 x 38.73 = 2.32 MN. (Key point: Roll force)

4. The torque per roll uses which arm, and total power uses how many rolls?
   A. Arm L; one roll
   B. Arm L/2; both rolls (P = 2 T omega)
   C. Arm R; one roll
   D. Arm L/2; one roll
   Answer: B. The resultant acts at mid-contact so T = F L/2 = 45 kN m, and P = 2 T omega = 942 kW. (Key point: Torque and power)

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