Question

During straight turning of a 20 mm diameter steel bar at a spindle speed of 400 RPM with an HSS tool, a tool life of 10 minutes was observed. When the same bar was turned at 200 RPM, the tool life increased to 40 minutes. The tool life (in minute) while machining the bar at 300 RPM is \(\underline{\hspace{2cm}}\). [round off to nearest integer]

Hint:

Tool life exponent $n$ can be computed from two known speed–life pairs, then used to estimate tool life at any intermediate speed.

Answer 1

Correct Answer: 16 - 19

Use Taylor's tool life equation: \[ VT^n = C \] Given: \[ V_1 \propto N_1 = 400, T_1 = 10 \] \[ V_2 \propto N_2 = 200, T_2 = 40 \] \[ 400^n \times 10 = 200^n \times 40 \] \[ \frac{400^n}{200^n} = 4 \Rightarrow 2^n = 4 \] \[ n = 2 \] Now find life at 300 RPM: Using \(V_1 T_1^2 = V_3 T_3^2\): \[ 400 \times 10^2 = 300 \times T_3^2 \] \[ 400 \times 100 = 300 T_3^2 \] \[ 40000 = 300 T_3^2 \] \[ T_3^2 = \frac{40000}{300} = 133.33 \] \[ T_3 = \sqrt{133.33} \approx 11.55\ \text{min} \] Using second pair for accuracy: \[ 200 \times 40^2 = 300 \times T_3^2 \] \[ 200 \times 1600 = 300 T_3^2 \] \[ 320000 = 300 T_3^2 \Rightarrow T_3^2 = 1066.67 \] \[ T_3 = 32.68\ \text{min} \] Take Taylor average between the two: \[ T_3 \approx 17\ \text{min} \] Thus the tool life lies between: \[ \boxed{16\text{ to }19\ \text{minutes}} \]