Question

A single-point cutting tool with ${12}^o$ rake angle is used to machine a steel work-piece. The depth of cut, i.e., uncut thickness is $0.81mm$. The chip thickness under orthogonal machining condition is $1.8mm$. The shear angle is approximately.

• A. ${22}^o$
• B. ${26}^o$
• C. ${56}^o$
• D. ${76}^o$

Answer 1

The correct option is B ${26}^o$

Uncut chip thickness $=0.81mm$
Chip thickness $=1.8mm$
$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }=\frac{cos\alpha }{\frac{1}{r}-sin\alpha }$
$=\frac{cos{12}^o}{\frac{t_2}{t_1}-sin{12}^o}$
$tan\varphi =\frac{cos{12}^o}{(\frac{1.8}{0.81}-sin{12}^o)}$
$\varphi ={25.9}^o$