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Derive an expression for cutting speed and tool life for minimum cost of production.
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Tool life:

  • Tool life based upon the criterion of the volume of material removed is derived as follows :

  • Volume of material removed per minute, $$ \begin{array}{l} \qquad \begin{array}{l} \forall=\pi \mathrm{D} \cdot \mathrm{t} \cdot \mathrm{f} \cdot \mathrm{N} \mathrm{mm}^{3} / \mathrm{min} \\ \forall=\pi \mathrm{D} \cdot \mathrm{t} \cdot \mathrm{f} \cdot \mathrm{N} \cdot \mathrm{T} \mathrm{mm}^{3} \end{array} \\ \text { Were, } \\ \qquad \begin{array}{l} \text { D=Diameter of workpiece, } \mathrm{mm} \\ \text { t= depth of cut, } \mathrm{mm} \\ \text { F=feed, } \mathrm{mm} / \mathrm{rev} \\ \text { N=revolution of work piece, } \mathrm{rpm} \\ \mathrm{T}=\text { time for tool failure, } \mathrm{min} \end{array} \end{array} $$

  • Cutting speed $$ \begin{aligned} \mathrm{V} &=[\pi \mathrm{D} \mathrm{N}] / 60 \mathrm{~m} / \mathrm{s} \\ &=[\pi \mathrm{D} \mathrm{N}] / 1000 \mathrm{~mm} / \mathrm{min} \\ \pi \mathrm{D} \mathrm{N} &=1000 \mathrm{~V} \end{aligned} $$

  • Tool life $\left[T_{L}\right]$ or volume of material removed, $$ \forall=1000 \mathrm{~V} \cdot \mathrm{t} \cdot \mathrm{f} \cdot \mathrm{T} \mathrm{mm}^{3} $$

Taylor tool life equation: - Cutting speed is the most important parameter which influence the tool wear and hence tool life

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  • The resultant relationship is a straight-line expressed in equation form called, Taylor tool life equation given by, $\mathrm{V} \mathrm{T}^{\mathrm{n}}=\mathrm{C}$ $\mathrm{V}=$ Cutting speed, $\mathrm{m} / \mathrm{min}$ $\mathrm{T}=$ Tool life, min $\mathrm{n}=$ Tool life index $\mathrm{C}=$ Machining index
  • Modified version is, $V T^{n} \cdot f^{n 1} \cdot d^{n 2}=C$ $f=$ feed rate, $\mathrm{mm} / \mathrm{rev}$

$\mathrm{d}=$ depth of cut, $\mathrm{mm}$

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