Question

A bar of 75 mm diameter is machined to reduce its diameter by orthogonal cutting. If the length of the cut chip is 73.5 mm and rake angle is ${15}^{\circ }$, the shear plane angle is

• A. ${16}^{\circ }$
• B. ${18}^{\circ }$
• C. ${23}^{\circ }$
• D. ${29}^{\circ }$

Answer 1

The correct option is B ${18}^{\circ }$

$l_c=73.5mm$

$\alpha ={15}^{\circ }$

$r=\frac{l_c}{l}=\frac{73.5}{\pi d}=\frac{73.5}{\pi \times 75}=0.31$

$\varphi =ta{n}^{-1}\left(\frac{rcos\alpha }{1-rsin\alpha }\right)$

$=ta{n}^{-1}\left(\frac{0.31cos15}{1-0.31sin15}\right)$

$\varphi ={18.03}^{\circ }$