Question

During an orthogonal machining operating on mild steel, the following results are obtained:
Uncut chip thickness = 0.3 mm
Chip thickness = 0.9 mm
Rake angle = 0°
Cutting force = 900 N
Thrust force = 400 N
Ultimate shear stress of the work materials = 380 $N/m{m}^2$
The minimum shear plane area is

• A. 1.914 $m{m}^2$
• B. 2.628 $m{m}^2$
• C. 3.828 $m{m}^2$
• D. 4.029 $m{m}^2$

Answer 1

The correct option is A 1.914 $m{m}^2$

$t=0.3mm;t_c=0.9mm;\alpha =0{}^{\circ };F_c=900N,F_T=400N;{\tau }_S=380N/m{m}^2$

$\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }=r=\frac{t}{t_c}=\frac{1}{3}$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{1}{3}\right)=18.43{}^{\circ }$

Now, the shear force $\left(F_s\right)$ is given by

$F_s=F_c\cos \varphi -F_T\sin \varphi =900\cos \left(18.43{}^{\circ }\right)-400\sin \left(18.43{}^{\circ }\right)$

$F_s=727.38N$

$\therefore$ Area of shear plane is,

$A_s=\frac{F_s}{{\tau }_s}=\frac{727.38}{380}=1.914m{m}^2$