Question

) A 30-mm-thick plate made of low carbon steel is to be reduced to 25 mm in one pass in a rolling operation. As the thickness is reduced, the plate widens by 4%. Yield strength of the steel = 174 MPa, and tensile strength = 450 MPa. The entrance speed of the plate = 77 m/min. Roll radius = 350 mm, and rotational speed = 45 rev/min. Determine (a) the minimum required coefficient of friction that would make this rolling operation possible, (b) exit velocity of the plate, and (c) forward slip.

Answer 1

Answer:

a. 0.12, b. 88.84m/min, c. -0.101

Explanation:

a. μf

Maximum depth of reduction d = μ²R where R is the Roll Radius

dm - dt = μ²Rr

30 - 25 = μ² x 350mm

μf = √(5/350) = 0.1195 = 0.12

b. Exit Velocity

since the volume of the plate would remain the same after the operation we can state that

Thickness initial x Width initial x Entrance Velocity = Thickness Final x Wdith Final x Exit Velocity

The plate widens 4% which means Wf = 1.04Wi

Substituting the values into above equation

30 x Wi x 77 = 25 x 1.04Wi x Vf

Vf = (30 x 77)/(25 x 1.04) = 88.84m/min

c. Forward Slip S

S = (Vf - Vr)/Vr

Vr = πDN where D is the diameter of the roller, and N is the Rotational Speed of the roller

Vr = π x (2 x 350 x 10^-3) x 45 = 98.91 m/min

S = (88.84 - 98.91)/98.91 = -0.101