Question

A mild steel plate has to be rolled in one pass such that the final plate thickness is 2/3^rd of the initial thickness, with the entrance speed of 10 m/min and roll diameter of 500 mm. If the plate widens by 2% during rolling, the exit velocity (in m/min) is _______

Answer 1

Concept:

Rolling:

• Rolling is the process in which the metals and alloys are plastically deformed into semi-finished or finished condition.
• This is done by passing the metals and alloys between circular rotating cylinders (rolls).
• The metal is drawn into the opening between the rolls by frictional forces between the metal and the roll surface.
• In deforming metal between rolls, the workpiece is subjected to high compressive force from the squeezing action of the rolls.
• There is no change in metal volume at a given point per unit time throughout the process.

B_iH_iV_i = B_fH_fV_f

where, B = Width, H = Thickness, V = Velocity

Calculation:

Let,

_Hi = Thickness of plate before rolling

_Hf = Thickness of plate after rolling

_Bi = Width of the plate before rolling

B_{f =} Width of the plate after rolling

_Vi = Entrance velocity = 10 m/min

_Vf = Exit velocity

\({{\rm{H}}_{\rm{f}}} = \frac{2}{3}{{\rm{H}}_{\rm{i}}}\), Bf = 1.02 Bi

As there is no change in metal volume at a given point per unit time throughout the process.

_HiBiVi = H_fBfVf

\(\Rightarrow {{\rm{H}}_{\rm{i}}}{{\rm{B}}_{\rm{i}}}{{\rm{V}}_{\rm{i}}} = \frac{2}{3}{{\rm{H}}_{\rm{i}}} \times 1.02{{\rm{B}}_{\rm{i}}} \times {{\rm{V}}_{\rm{f}}}\)

⇒ \({V_f} = \frac{{{V_i}}}{{0.68}} = \frac{{10}}{{0.68}} = 14.7\;m/min\)