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Manufacturing Engineering
Machining Processes
GATE 2017
Expert

During the turning of a 20 mm-diameter steel bar at a spindle speed of 400 rpm, a tool life of 20 minute is obtained. When the same bar is turned at 200 rpm, the tool life becomes 60 minute. Assume that Taylor's tool life equation is valid. When the bar is turned at 300 rpm, the tool life (in minute) is approximately

  • A25
  • 32
  • C40
  • D50

Explanation

V=πDN/1000. At 400 rpm: V1=25.13 m/min, T1=20 min. At 200 rpm: V2=12.57 m/min, T2=60 min. From VTⁿ=C: (25.13/12.57)=(60/20)ⁿ → 2=3ⁿ → n=ln2/ln3≈0.631. C=25.13×20^0.631≈166.3. At 300 rpm: V3=18.85 m/min. T3=(C/V3)^(1/n)=(166.3/18.85)^1.585≈8.83^1.585≈31.6 min ≈ 32 min.

Source: GATE ME 2017 Session 2 (ME-2) (2017)

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