Question

In the turning experiment, a 50 mm diameter bar is turned using a tool having a rake angle $6^{\circ }$ with orthogonal machining operation at a cutting speed of 40 m/min, chip flow velocity is 20 m/min. The feed and depth of cut are 0.24 mm/rev and 1.8 mm respectively. $F_c$ is observed as 800 N and $F_t$ as 290 N. The shear stress induced in the work material during machining operation is

• A. 19.925
• B. 27.68
• C. 573.77
• D. 616.98

Answer 1

The correct option is D 616.98

Given data:
$D=50\text{mm,}f=0.24\text{mm/rev,}\alpha =6^{\circ }$
$V_c=40\text{m/min,}{V}_f=20\text{m/min,}$
$d=1.8\text{mm}$
$F_c=800\text{N}$
$F_t=290\text{N}$
$\text{Shear stress}\left(\tau \right)=\frac{F_s}{L_s\times b}=\frac{F_s\sin \varphi }{t_1\times b}$

For turning and boring operation
$t_1=f\cos C_s=f\sin \lambda =f=0.24$
$b=\frac{d}{\sin \lambda }=d=1.8$

$F_s=F_c\cos \varphi -F_t\sin \varphi$

$r=\frac{V_f}{V_c}=\frac{20}{40}=0.5$

$\varphi ={\tan }^{-1}\left(\frac{r\cos \alpha }{1-\sin \alpha }\right)=27.68$

$F_s=800\cos {27.68}^{\circ }-290\sin {27.68}^{\circ }$
$=573.73\text{N}$

${\tau }_{\text{ultimate}}=\frac{573.73\sin {27.68}^{\circ }}{f\times d}=616.93$