Question:

Cylindrical workpieces of diameter 60 mm and length 400 mm are machined on a lathe at a cutting speed of 25 m/min and a feed of 0.2 mm/rev. The Taylor’s tool life parameters \(C\) and \(n\) for this setup are 75 and 0.25, respectively. The tool changing time is 3 minutes. With a labor and overhead cost of ₹5 per minute, the tool changing cost per piece is ₹_________ (rounded off to 2 decimal places).

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To calculate tool changing cost, use the cutting time formula and then add the tool changing time. Multiply by the labor and overhead cost to get the total cost per piece.
Updated On: June 02, 2025
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Solution and Explanation

Given:
Diameter, \( D = 60 \, \text{mm} \)
Length, \( L = 400 \, \text{mm} \)
Cutting speed, \( V = 25 \, \text{m/min} \)
Feed, \( f = 0.2 \, \text{mm/rev} \)
Taylor's tool life constants: \( C = 75 \), \( n = 0.25 \)
Tool changing time, \( t_c = 3 \, \text{min} \)
Labor and overhead cost = ₹5/min

Step 1: Tool Life using Taylor's Equation
\[ VT^n = C \quad \Rightarrow \quad T = \left( \frac{C}{V} \right)^{\frac{1}{n}} = \left( \frac{75}{25} \right)^{\frac{1}{0.25}} = (3)^4 = 81 \, \text{min} \]

Step 2: Machining Time per Piece
\[ N = \frac{1000 \cdot V}{\pi \cdot D} = \frac{1000 \cdot 25}{\pi \cdot 60} \approx 132.63 \, \text{rev/min} \]
\[ t_m = \frac{L}{f \cdot N} = \frac{400}{0.2 \cdot 132.63} \approx \frac{400}{26.526} \approx 15.08 \, \text{min} \]

Step 3: Number of Pieces per Tool Life
\[ \text{Pieces per tool} = \frac{T}{t_m} = \frac{81}{15.08} \approx 5.37 \]

Step 4: Tool Changing Cost per Piece
\[ \text{Cost per change} = t_c \times \text{Cost/min} = 3 \times 5 = ₹15 \]
\[ \text{Cost per piece} = \frac{15}{5.37} \approx ₹2.79 \]

Final Answer: ₹2.79
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