Discussion
In a single pass rolling process using 410 mm diameter steel rollers, a strip of width 140 mm and thickness 8 mm undergoes 10 % reduction of thickness. The angle of bite in radians is
Answer: C
The angle of bite is about 0.062 rad.
A 10 % reduction of 8 mm is a draft of 0.8 mm, and the roll radius is 205 mm. The contact angle satisfies:
\[ \begin{aligned} \cos\alpha &= 1-\frac{\Delta h}{2R} = 1-\frac{0.8}{410} = 0.99805 \\ \alpha &= 0.0625\ \text{rad} \end{aligned} \]
The small-angle form \(\alpha \approx \sqrt{\Delta h/R} = \sqrt{0.8/205} = 0.0625\) rad gives the same result.
0.031 is half the correct value and 0.006 and 0.600 are the right digits with the decimal point misplaced. The strip width plays no part in the bite angle.
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