Question

Common Data:
In an orthogonal machining operation:
Uncut thickness $=0.5mm$
Cutting speed $=20m/min$
Rake angle $={15}^o$
Width of cut $=5mm$
Chip thickness $=0.7mm$
Thrust force $=200N$
Cutting force $=1200N$
Assume Merchant's theory.

The Values of shear angle and shear strain, respectively, are

• A. ${30.3}^o$ and $1.98$
• B. ${30.3}^o$ and $4.23$
• C. ${40.2}^o$ and $2.97$
• D. ${40.2}^o$ and $1.65$

Answer 1

The correct option is D ${40.2}^o$ and $1.65$

Given;
$\text{Uncut thickness,}{t}_1=0.5mm$
$\text{Cutting speed, v}=20m/min$
$\text{Rake angle,}\alpha ={15}^o$
$\text{Width of cut,}w=5mm$
$\text{Chip thickness,}{t}_2=07mm$
$\text{Thrust force,}{F}_T=200N$
$\text{Cutting force,}{F}_C=1200N$

Chip thickness ratio or Cutting Ratio:
$r=\frac{t_1}{t_2}=\frac{0.5}{0.7}=0.714$

Shear plane angle:
$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }=\frac{0.714cos{15}^o}{1-0.714sin{15}^o}$
$\therefore \varphi =ta{n}^{-1}\left(0.847\right)={40.2}^o$

Shear strain:
$\gamma =cot\varphi +tan(\varphi -\alpha )$
$=cot{40.2}^o+tan({40.2}^o-{15}^o)$
$=1.181+0.4716$
$=1.65$