GATE 2017 ME (ME1) – Question 65

Machining and Machine Tool Operations · Machining operations, machine tools, cutting tools and tool geometry · 2 marks · Numerical answer

Two tools obey Taylor tool-life equations: Tool 1, $VT^{0.1}=150$; Tool 2, $VT^{0.3}=300$. V is cutting speed in m/min and T is life in minutes. Find the break-even speed, beyond which Tool 2 has longer life, in m/min.

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Correct answer: 105 to 107

Explanation

At break-even both tools have the same V and T. Dividing the second equation by the first gives $T^{0.2}=2$, so $T=2^5=32$ min. Then $V=150/32^{0.1}=106.066$ m/min. Above this speed, Tool 1’s life falls more rapidly because its life varies as $V^{-10}$, compared with $V^{-10/3}$ for Tool 2. Thus the crossing speed is about 106.07 m/min.

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