Question

During orthogonal metal cutting process, the following data is obtained:
Rake angle = 12°
Chip thickness ratio = 0.5
Undeformed thickness = 0.7 mm
Cutting speed = 2.4 m/s
If the mean thickness of primary shear zone is 25 microns, then shear strain rate ($in{s}^{-1}$) during the process is

• A. 4.9 × ${10}^4$
• B. 0.98 × ${10}^4$
• C. 9.8 × ${10}^4$
• D. 0.49 × ${10}^4$

Answer 1

The correct option is C 9.8 × ${10}^4$

$r=0.5;\alpha =12{}^{\circ };v=2.4m/s$

Now, $\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }=\frac{0.5\cos 12{}^{\circ }}{1-0.5\sin 12{}^{\circ }}$

$\varphi =ta{n}^{-1}\left(0.5458\right)=28.63{}^{\circ }$

Also, shear velocity is,

$v_s=\frac{v\cos \alpha }{\cos (\varphi -\alpha )}=\frac{2.4\cos 12{}^{\circ }}{\cos (28.63-12)}=2.45m/s$

$\therefore \text{Shear strain rate}=\frac{v_s}{t_s}=\frac{2.45}{25\times {10}^{-6}}=9.8\times {10}^4{s}^{-1}$