Question

A 1000 mm wide steel sheet made of C35 is normalized by cold rolling 10 mm thick

deformed to 5 mm. The rollers, 600 mm in diameter, run at a peripheral speed of 0.12 m/s.
The deformation efficiency is 55%.
Find out:
a) the roller force
b) the roller torque
c) the performance on the pair of rollers.

Answer 1

a. 20.265 MN

b. 0.555 MNm

c. 403.44 KW

Step-by-step explanation:

Given:-

- The width ( w ) = 1000 mm

- Original thickness ( to ) = 10 mm

- Final thickness ( t ) = 5 mm

- The radius of the rollers ( R ) = 600 mm

- The peripheral speed of the roller ( v ) = 0.12

- Deformation efficiency ( ε ) = 55%

Find:-

a) the roller force ( F )

b) the roller torque ( T )

c) the performance on the pair of rollers. ( P )

Solution:-

- The process of flat rolling entails a pair of compressive forces ( F ) exerted by the rollers on the steel sheet that permanently deforms.

- The permanent deformation of sheet metal is seen as reduced thickness.

- We will assume that the compressive force ( F ) acts normal to the point of contact between rollers and metal sheet.

- The roll force ( F ) is defined as:

`F =L*w*Y_a_v_g`

Where,

L: The projected length of strip under compression

Y_avg: The yielding stress of the material = 370 MPa

- The projected length of strip under compression is approximated by the following relation:

`L = √(R*( t_o - t_f )) \\\\L = √(0.6*( 0.01 - 0.005 )) \\\\L = 0.05477 m`

- The Roll force ( F ) can be determined as follows:

`F = (0.05477)*(1 )*(370*10^6 )\\\\F = 20.265 MN`

- The roll torque ( T ) is given by the following relation as follows:

`T = (L)/(2) * F\\\\T = (0.05477)/(2) * 20.265\\\\T = 0.555 MNm`

- The rotational speed of the rollers ( N ) is determined by the following procedure:

`f = (v)/(2\pi* R) = (0.12)/(2*\pi 0.6) = 0.03181818 (rev)/(s) \\\\N = f*60 = 1.9090 rpm`

- The power consumed by the pair of rollers ( P ) is given by:

`P = (2\pi * F * L * N)/(e*60,000) KW \\\\P = (2\pi * ( 20.265*10^6) * (0.05477) * (1.90909 ) )/(60,000*0.55) KW\\\\P = 403.44 KW`