Question

In an orthogonal cutting operation on a work piece of width 3 mm, the uncut chip thickness was 0.5 mm and the tool rake angle was $2^{\circ }$. It was observed that the chip thickness was 1.25 mm and cutting force was measured to be 900 N and the thrust force was found to be 500 N. The shear strength of work material is

• A. 161.7 MPa
• B. 64.7 MPa
• C. 388 MPa
• D. 398.5 MPa

Answer 1

The correct option is A 161.7 MPa

Given data:
$b=3\text{mm,}{t}_1=0.5\text{mm}$
$\alpha =2^{\circ },t_2=1.25\text{mm}$

$F_c=900\text{N,}{F}_T=500\text{N}$
$r=\frac{t_1}{t_2}=\frac{0.5}{1.25}=0.4$

$\varphi ={\tan }^{-1}\left(\frac{r\cos \alpha }{1-r\sin \alpha }\right)={22}^{\circ }$

$\text{Shear strength}\left(\tau \right)=\frac{Fs}{\frac{bt}{\sin \varphi }}$

$F_s=F_c\cos \varphi -F_t\sin \varphi$
$=900\cos 22-500\sin 22=647.16\text{N}$

$\tau =\frac{647.16}{\frac{3\times 0.5}{\sin 22}}{\text{N/mm}}^2$

$=161.62=161.7\text{MPa}$