Question

The tool life test was conducted using disposable carbide insert tools and medium carbon steel work pieces. Following results were obtained.

Sl. No.	Cutting speed (m/min)	Tool life (min)
1.	35	160
2.	70	40

An insert has 4 cutting edges, insert replacement time is 1min and hourly machining rate is Rs. 1200. If the cost of insert is Rs. 500, the cutting speed for minimum cost of machining is

• A. 164.4 m/min
• B. 1371.7 m/min
• C. 88.544 m/min
• D. 11.33 m/min

Answer 1

The correct option is A 164.4 m/min

$V_1{T}_1^n=V_2{T}_2^n$

$35\times (160{)}^n=70(40{)}^n$

$(40{)}^n=2$

$nln4=ln2$

$n=0.5$

$C=442.71$

Tooling cost,

$C_g=\frac{500}{4}=125/-$

$T_C=1\text{min}$

$L_m=\frac{1200}{60}\text{Rs/min}=20\text{Rs/min}$

$T_0=(T_c+\frac{C_g}{L_m})\left(\frac{1-n}{n}\right)$

$=(1+\frac{125}{20})\left(\frac{1-0.5}{0.5}\right)=7.25$

$V_0{T}_0^n=C\Rightarrow V_0\times {7.25}^{0.5}=442.71$

$V_0=164.41\text{m/min}$

$V_0=\text{Optimum speed for minimum cost)}$

$C_m=\frac{\pi DL}{1000fV}\times L_m$

$=\frac{\pi \times 100\times 300\times 20}{1000\times 0.25\times 177}$

$=42.59=42.6$

$C_t=\frac{\pi DL{V}^{\frac{1}{n}-1}}{1000f{C}^{\frac{1}{n}}}.C_g$

$\frac{\pi \times 100\times 300\times {177}^{\frac{1}{0.5}1}}{1000\times 0.25\times (442.71{)}^{\frac{1}{0.5}}}\times 125$

$=42.55$

$\text{Total variable cost}=C_m+C_t$

$42.59+42.55$

$=85.147=85.2/-$