Question

Linked Data:
During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle, the following data is obtained:
Uncut chip thickness $=0.25mm$
Chip thickness $=0.75mm$
Normal force $=950N$
thrust force $=475N$

The shear angle and shear force, respectively, are

• A. ${71.565}^o,150.12N$
• B. ${9.218}^o,861.64N$
• C. ${18.435}^o,751.04N$
• D. ${23.157}^o,686.66N$

Answer 1

The correct option is C ${18.435}^o,751.04N$

Given:
$\alpha =0^o,t_1=0.25mm,t_2=0.75mm$
$w=b=2.5mm,F_c=950N,F_T=475N$

Chip thickness ratio:
$r=\frac{t_1}{t_2}=\frac{0.25}{0.75}=\frac{1}{3}=0.333$

Shear plane angle for zero rake angle:
$\varphi =ta{n}^{-1}\left(r\right)=ta{n}^{-1}\left(0.333\right)={18.43}^o$

Friction angle for zero rake angle:
$\beta =ta{n}^{-1}\left(\frac{F_T}{F_C}\right)=ta{n}^{-1}0.5={26.56}^o$

Shear force:
$F_s=\frac{F_C}{cos(\beta -\alpha )}cos(\varphi +\beta -\alpha )$

$=\frac{950}{cos{26.56}^o}cos\left({45}^o\right)=751N$