Question

The following data are obtained during orthogonal machining with Merchants theory.

Ultimate shear strength of work material = 600 MPa,
Rake angle of the tool $=6^{\circ }$,
Coefficient of friction at chip tool interface $=0.2$,
Feed and depth of cut = 0.2 mm/rev and 3 mm respectively. Taylor's tool life equation is $V.T^{0.4}=100$.

The power required during machining operation is

• A. 414 W
• B. 572 W
• C. 2117 W
• D. 2932 W

Answer 1

The correct option is A 414 W

$\mu =0.2,\alpha =6^{\circ },$

$n=0.4,C=100$

${\tau }_{\text{ultimate}}=600\text{MPa}$

$f=0.2\text{mm/rev}$

$d=3\text{mm}$

$V_{opt}=C{\left[\frac{n}{1-n}\frac{L_m}{C_g}\right]}^n$

$=100{\left[\frac{0.4}{0.6}\frac{\frac{75}{60}}{15}\right]}^{0.4}$

$=31.46\text{m/min}=31.5\text{m/min}$

$t_1=0.2\text{mm,}b=3\text{mm,}$

$\varphi =\frac{{90}^{\circ }+\alpha -\beta }{2}$

$\varphi =\frac{{90}^{\circ }+6^{\circ }-{11.3}^{\circ }}{2}={42.34}^{\circ }$

$600{\text{N/mm}}^2=\frac{F_s}{\frac{t_1}{\sin \varphi }\times b}$

$F_s=534.5\text{N}$

$F_s=\frac{F_c}{\cos (\beta -\alpha )}\times \cos (\varphi +\beta -\alpha )$

$\frac{534.5\times \cos (11.3-6)}{\cos (42.34+11.3-6)}=F_c$

$\Rightarrow F_c=789.88\text{N}\approx 790\text{N}$

$P=F_c\times V_c$

$W=790\times \frac{31.5}{60}$

$P=W=414.68W$