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Fundamental Tricks to Solve Numericals

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Question

The following data relate to an orthogonal turning process: back rake angle $= 15 d e g .$ width of cut $= 2 m m$ chip thickness $= 0.4 m m$ feed rate $= 0.2 m m / r e v$

The shear angle is

A

14^{0}

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B

25^{0}

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C

29^{0}

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D

75^{0}

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Solution

The correct option is C $29^{0}$
$α = 15^{0}, b = 2 m m, t_{2} = 0.4 m m$
$t_{1} = 0.2 m m$
$r = \frac{t_{1}}{t_{2}} = \frac{0.2}{0.4} = \frac{1}{2} = 0.5$
$ϕ = t a n - 1 (\frac{0.5 c o s 15^{0}}{1 - 0.5 s i n 15^{0}}) = 29^{0}$

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