Question

In orthogonal turning of a low carbon steel bar of diameter $150mm$ uncoated carbide tool the cutting velocity is $90m/min$. The feed is $0.24mm/rev$ and the depth of cut is $2mm$. The chip thickness obtained is $0.48mm$. If the orthogonal rake angle is zero and the principal cutting edge angle is ${90}^o$, the shear angle in degree is

• A. $20.56$
• B. $26.56$
• C. $30.56$
• D. $36.56$

Answer 1

The correct option is B $26.56$

Given that,
Feed rate: $f=0.24mm/rev$
Chip thickness: $t_2=0.48mm$
$\text{Principal cutting edge angle},\Psi ={90}^o$
$\text{Rake angle}:\alpha =0$

We know that
Uncut chip thickness,
$t_1=fsin\Psi$
$=0.24sin{90}^o$
$t_1=0.24mm$

Cutting ratio: $r=\frac{t_1}{t_2}=\frac{0.24}{0.48}=0.5$
$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }$
$=\frac{0.5cos{0}^o}{1-0.5sin{0}^o}=0.5$

$\text{Shear angle}:\varphi =ta{n}^{-1}\left(0.5\right)={26.56}^o$