In a rolling operation, the thickness of a rectangular piece is reduced from t₁ to t₂. The radius of both the rolls are equal to R. The angle of bite (θ) will be given by:

Explanation:

t₁ = thickness of bar before rolling

t₂ = thickness of bar after rolling

R = Radius of Rollers

θ = ∠ AOC = Bite angle

Bite Angle:

It is the angle subtended at the centre by the contact arc of roll-work.

Here BC = B’C’

∴ t₁ – t₂ = BC + B’C’ ⇒ 2BC

\( \Rightarrow BC = \frac{{{t_1} - {t_2}}}{2}\)

\({\rm{cosθ }} = \frac{{{\rm{OB}}}}{{{\rm{OA}}}}\)

\(\therefore {\rm{cosθ }} = \frac{{{\rm{OC}} - {\rm{BC}}}}{{{\rm{OA}}}}\)

\( \Rightarrow {\rm{cosθ }} = \frac{{R - \left( {\frac{{{t_1} - {t_2}}}{2}} \right)}}{R}\)

\( \Rightarrow {\rm{cosθ }} = 1 - \frac{{{t_1} - {t_2}}}{{2R}}{\rm{\;}}\)

\( \Rightarrow θ = {\cos ^{ - 1}}\left[ {1 - \frac{{{t_1} - {t_2}}}{{2R}}} \right]\)

Alternate method:

t₁ - t₂ = 2R (1 - cosθ)  

\(\frac{t_1 - t_2}{2R}=1 - cos \theta\)

\( \Rightarrow {\rm{cosθ }} = 1 - \frac{{{t_1} - {t_2}}}{{2R}}{\rm{\;}}\)

\( \Rightarrow θ = {\cos ^{ - 1}}\left[ {1 - \frac{{{t_1} - {t_2}}}{{2R}}} \right]\)