Question

In an orthogonal cutting test with a tool of rake angle 15 deg, the following observations were made: chip thickness ratio = 0.37, horizontal cutting force = 1000 N, vertical cutting force = 1500 N.

The machining constant for the given machining operation is

• A. ${76.82}^{\circ }$
• B. ${65}^{\circ }$
• C. ${55.8}^{\circ }$
• D. ${49}^{\circ }$

Answer 1

The correct option is B ${65}^{\circ }$

$2\varphi +\beta -\alpha ={\cot }^{-1}\left(k\right)$

where k is the machining constant

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

$={\tan }^{-1}\left(\frac{0.37\cos 15}{1-0.37\sin 15}\right)$

$={21.56}^{\circ }$

$\beta ={\tan }^{-1}\left(\mu \right)={\tan }^{-1}\left(0.75\right)$

$\beta ={36.86}^{\circ }$

$k=\cot (2\varphi +\beta -\alpha )=\cot (64.98\approx \cot {65}^{\circ }$

$\text{As examiner has asked in degrees,}{65}^{\circ }$