In a orthogonal cutting set up, depth of cut was 10mm, feed=1mm/rev cutting speed is 60 rpm, back rake angle =10°, chip thickness ratio is 0.33, shear stress of material is at zero compressive stress is 1000 kg/sq. cm. Assume the value constant K in equation 2 $\varphi+\beta–\alpha=cot^{-1}k$, is 0.2.

Calculate the resultant force, rate of metal removal, shear strain, HP at the tool per cubic cm of metal removal /minute.

depth of cut; $b_1$=10mm

f=1mm/rev

chip thickness ratio;$ r_c$=0.33

As per A.S.A. system

$\alpha=10^0$

$f_{so}=1000kg/m^2$

=$98.1 N/mm^2$

k=0.2

$2\varphi+\beta –\alpha= cot^{-1}k=C_m$

cot $C_m= 0.2$

therefore $C_m$ = 78.69

$\therefore 2\varphi+\beta –\alpha $= 78.69……………….1

also,

$tan \Phi = \frac{r_c \times cos \alpha}{1 - r_c sin a} \Rightarrow tan \Phi= \frac{0.33\times cos …