Depth of cut is 6 $\mathrm{mm}$ and the rate of feed 0.2 $\mathrm{mm} / \mathrm{rev}$ .

Two types of tools are available for this purpose:

(i) H.S.S.

(ii) Tungsten Carbide.

The following are the data available:

HSS costs an average of Rs. 30 per edge and carbide costs Rs. 75 per edge. Take the operating cost as Rs. 120 per hour. Analyze the selection of the tool material based on minimum cost of machining. $\mathrm{MU}-\mathrm{Dec}, 2012,10 \mathrm{Marks}$

Solution: $\mathrm{d}=50 \mathrm{mm}, l=300 \mathrm{mm},$ depth of cut $=6 \mathrm{mm}, \mathrm{f}=0.2 \mathrm{mm} / \mathrm{rev}$

For H.S.S tool:

$\mathrm{T}_{1}=20 \mathrm{mm}, \mathrm{V}_{1}=40 \mathrm{mm}, \quad \mathrm{t}_{d}=3 \mathrm{mm}$ $\mathrm{T}_{2}=35 \mathrm{mm}, \quad \mathrm{V}_{2}=31$

$x=\frac{\log V_{1}-\log V_{2}}{\log T_{2}-\log T_{1}}=\frac{\log 40-\log 31}{\log 35-\log 20}$

$=\frac{1.6021-1.4914}{1.5441-1.3010}=0.4554$

$\mathrm{VT}^{\mathrm{n}}=40+20^{0.4554}=156.51$

$y=R_{e} 30$

$x=\frac{120}{60}=R s .2 / \mathrm{min}$

$\mathrm{R}=\mathrm{t}_{\mathrm{d}}+\frac{\mathrm{y}}{\mathrm{x}}=3+\frac{30}{2}=18$

$\mathrm{T}_{\mathrm{o}}=\mathrm{R}\left[\frac{1}{\mathrm{n}}-1\right]=18\left[\frac{1}{0.4554}-1\right]$

$=21.53 \mathrm{min}$

$\mathrm{V}_{\mathrm{o}}=\frac{156.51}{21.53^{0.4554}}=38.68 \mathrm{m} / \mathrm{min}$

$t_{c o}=\frac{300}{0.2 \times \frac{38.68 \times 1000}{\pi \times 50}}$

$=6.09 \mathrm{min}$

Machinery cost $=x\left[t_{e}+\frac{t c_{o}}{1-n}\right]$

$=2\left[0+\frac{6.09}{1-0.4554}\right]$

$=\quad \mathrm{Rs} .22 .365$ per piece

For Carbide tool

$\begin{array}{rlrl}{T_{1}} & {=} & {15,} & {V_{1}=125} \\ {T_{2}} & {=45,} & {} & {V_{2}=85}\end{array}$

$\mathrm{n}=\frac{\log 125-\log 85}{\log 45-\log 15}=\frac{2.0969-1.9294}{1.6532-1.1761}=0.351$

$\mathrm{VT}^{\mathrm{n}}=125 \times 15^{0.351}=323.38$

$y=\mathrm{R}_{\mathrm{S}} .75, \quad \mathrm{x}=\mathrm{Rs} .2, \quad \mathrm{t}_{\mathrm{d}}=3 \mathrm{mm}$

$\mathrm{R}=\mathrm{t}_{\mathrm{d}}+\frac{\mathrm{y}}{\mathrm{x}}=\left(3+\frac{75}{2}\right)=40.5$

$\mathrm{T}_{\mathrm{o}}=\mathrm{R}\left(\frac{1}{\mathrm{n}}-1\right)=40.5\left(\frac{1}{0.357}-1\right)$

$=74.88$ minutes

$\mathrm{V}_{\mathrm{o}}=\frac{323.38}{74.88^{0.351}}=71.09 \mathrm{m} / \mathrm{min}$

$t_{\mathrm{co}}=\frac{300}{0.2 \times \frac{71.09 \times 1000}{\pi \times 50}}=3314 \mathrm{m}$

$\operatorname{cost}=x\left[T_{c}+t_{c o}\left(1+\frac{R}{T}\right)\right]$

$=2\left[0+3.314\left(1+\frac{40.5}{74.88}\right)\right]$

$=\mathbf{R} s .10 .21$

Machinery which can cost almost half.