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 1 min 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

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

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

$(4{)}^n=2$

$nln4=ln2$

n = 0.5

C = 442.71

Tooling cost,

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

$T_C=1min$

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

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

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

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

$V_0=164.41m/min$

($V_0=Optimumspeedforminimumcost)$

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

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

= 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

Total variable cost = $C_m+C_t$

= 42.59 + 42.55

= 85.147 = 85.2/-