The optimal cutting parameters for the maximum rate of production are obtained by minimizing the total production time per part.

The total production time per part is given by

$T_p = t_ e + t_c + \frac{t_dt_c}{T} = t_e + t_c[1 + \frac{t_d{}_o}{T}]$

$te + \frac{\pi DL}{60000Vf}[1 + t_dC^{-\frac{1}{n}}V^{\frac{1}{n}}f^{-\frac{m}{n}}]$

Partial differentiation of this equation with with respect to V and equating to zero gives:

$\frac{t_d}{T'_o}(\frac{1}{n}-1) = 1$

or $T'_o = t_d(\frac{1}{n} - 1)$

At the maximum optimum feed $f_o$,the corresponding optimum cutting speed is given

$V'_o = \frac{C}{(T'_o)^n(f'_o)^m}$

For the same optimum feed

$\frac{V'_o}{V_o} = (\frac{T_o}{T'_o})^n = [\frac{R(\frac{1}{n} - 1)}{t_d(\frac{1}{n} - 1)}]^n$

$= (\frac{R}{t_d})^n$

$V'_o$ can be equal to $V_o$ only if R = $t_d$

i.e. $t_d + \frac{y}{x} = t_d$

Question

A 200 mm long and 30 mm bar is being turned on lathe with a feed of 0.25 mm/rev. The operating cost is $5 paise/min$ while cost is Rs 1 per cutting edge. Tool changing time is 1 min for each cutting edge. Compare the machining cost per component while operating under most economical condition for following materials :

Solution :

$l= 200 mm$

$d = 30 mm$

$f= 0.25 mm/rev$

$x = 0.5 min$,

y = $R_e$ 1

$f_d = 1 min$

For $X VT^{0.1} = 67$ For $Y VT^{0.1} = 90$

$VT^{0.1} = c$

$n = 0.1$

$R = f_d + \frac{y}{x} = 1 + \frac{1}{0.05} = 21$

$T_o = R[\frac{1}{x} - 1]$ = $21[\frac{1}{0.1}] = 189 min$

For material X

$V_o = \frac{67}{T_o^{0.1}} = \frac{67}{189^{0.1}} = 39.67 m/min$

$FC_o = \frac{200}{0.25 \times \frac{39.67 \times 1000}{x \times 30}} = 1.9 mm$

$C_o = x[t_e + \frac{tc_o}{1-n}]$ without material cost

$= 5[0+ \frac{1.9}{0.9}] = 10.55 paise$

Wher $C_1$ is the machinery cost /price in paise

For Material Y

$V_oy_o = \frac{90}{189^{0.1}} = 53.28 m/min$

$t_c{o} = \frac{200}{0.25 \times \frac{53.28 \times 1000}{\pi \times 30}} = 1.415 min$

Machinery cost = $T = 5[0 + \frac{1.415}{9}] = 7.86 paise$