In the tool wear plots of Figure 224 complete failure of the

In the tool wear plots of Figure 22.4, complete failure of the cutting tool is indicated by the end of each wear curve. Using complete failure as the criterion of tool life instead of 0.50 mm flank wear, the resulting data are: (a) v = 160 m/min, T = 5.75 min; (b) v = 130 m/min, T = 14.25 min; and (c) v = 100 m/min, T = 47 min. Using these results:

(1) Determine the parameters n and C in the Taylor tool life equation for this data.

(2) If the cutting speed v=80m/min, determine the tool life for this application.

Solution

We know the Taylor tool life equation as below-

VT^n = C

On taking log on both side,

ln V + n ln T = ln C ……………….(1)

(a) V = 160 m/min, T = 5.75 min;

(b) V = 130 m/min, T = 14.25 min; and

(c) V = 100 m/min, T = 47 min

Putting value of (a) in eq. (1)

ln 160 + n ln 5.75 = ln C

1.7492 n – ln C = - 5.075174 ……………….(2)

On putting (b)

ln 130 + n ln 14.25 = ln C

2.6567569 n – ln C = - 4.86753445……………….(3)

on solving 2 and 3 we get

ln C = 5.47537271 or C = 238.74

n = 0.228789566

Now check for (c) option

100 x (47^ 0.228789566) = C

C = 241 approx 238.74 hence it is correct.

If cutting speed V = 80

80 x (T) ^ 0.228789566 = 238.74

T = 119.1 min ……………………..Ans.