Question

A plate that is 250 mm wide and 25 mm thick is to be reduced in a single pass in a two-high rolling mill to a thickness of 20 mm. The roll has a radius = 500 mm, and its speed = 30 m/min. The work material has a strength coefficient = 240 MPa and a strain hardening exponent = 0.2. Determine (a) roll force, (b) roll torque, and (c) power required to accomplish this operation.

Answer 1

a) Roll force (Fr) = 1500 MPa.mm

(b) Roll torque (Tr) = 750,000 MPa.mm^2

(c) Power required (P) = 375,000,000 MPa.mm^3/s

To determine the roll force, roll torque, and power required for the rolling operation, we need to use the following formulas:

(a) Roll force (Fr):

Fr = (0.5 × width × thickness × strength coefficient) / roll radius

(b) Roll torque (Tr):

Tr = Fr × roll radius

(c) Power required (P):

P = Tr × rotational speed

Let's calculate each of these values step by step:

Given data:

Width (W) = 250 mm

Initial thickness (t1) = 25 mm

Final thickness (t2) = 20 mm

Roll radius (R) = 500 mm

Roll speed (V) = 30 m/min

Strength coefficient (K) = 240 MPa

Strain hardening exponent (n) = 0.2

(a) Roll force (Fr):

Fr = (0.5 × W × (t1 - t2) × K) / R

Substituting the given values:

Fr = (0.5 × 250 mm × (25 mm - 20 mm) × 240 MPa) / 500 mm

Fr = (0.5 × 250 mm × 5 mm × 240 MPa) / 500 mm

Fr = 1500 MPa.mm

(b) Roll torque (Tr):

Tr = Fr × R

Substituting the calculated roll force and roll radius:

Tr = 1500 MPa.mm × 500 mm

Tr = 750,000 MPa.mm^2

(c) Power required (P):

P = Tr × V

Converting the roll speed from m/min to mm/s:

V = 30 m/min × (1000 mm/1 m) / (60 s/1 min)

V = 500 mm/s

Substituting the calculated roll torque and roll speed:

P = 750,000 MPa.mm^2 × 500 mm/s

P = 375,000,000 MPa.mm^3/s

Note: The units for power are MPa.mm^3/s. If you prefer another unit, you can convert it accordingly (e.g., Watts, kilowatts).

So, the results are as follows:

(a) Roll force (Fr) = 1500 MPa.mm

(b) Roll torque (Tr) = 750,000 MPa.mm^2

(c) Power required (P) = 375,000,000 MPa.mm^3/s