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,

$tan \Phi = \frac{r_c \times cos \alpha}{1 - r_c sin a} \Rightarrow tan \Phi= \frac{0.33\times cos 10}{1-0.33 sin 10}$

$\therefore \varphi=19.02^0$

1) R

$F_s= R cos(\varphi+\beta –\alpha) =R (0.504)$

$F_N= R Sin(\varphi+\beta–\alpha) = R( 0.863)$

$f_s = f_{so} + k_1f_N$

$\therefore \frac{F_s}{\frac{t_1b_1}{sin \Phi}}= 98.1 +\frac{F_N}{\frac {t_1b_1}{sin \Phi}} \Rightarrow \frac{R (0.504)}{30.68}= 98.1 +\frac{R( 0.863)}{30.68}$

Therefore R=9432.69 N……….ans

2) MRR

volume of material removed=$t_1 \times b_1 \times V_c \times 1000 (V_c$=60m/min mistake in question)

MRR=$ 1 \times 10 \times 60 \times 1000$

MRR= $600 \times 10^3 mm^3/min$

3)Shear strain

Shear strain=$cot \varphi +tan(\varphi- \alpha)$

=cot (19.02) + tan(9.02)

shear strain= 3.05

4) HP at the tool per cubic cm of metal removal /minute.

From merchant circle

$cos (\beta-\alpha) =\frac{F_c}{R}$

therefore, $F_c=7156.6N$

$P_c=\frac{F_c(in kg) \times V_c}{4500}\Rightarrow Pc =\frac{729.59 \times 60}{4500}$

therefore, $P_c$=9.726HP

Now HP per $cm^3/min$

i.e. $\frac{9.726}{600}$

= 0.01621 HP per $cm^3/min$