Chip thickness and force in up milling (conventional): How does the cutting force vary per tooth engagement during one pass?

Difficulty: Easy

Correct Answer: Increases from zero to maximum

Explanation:


Introduction / Context:
Up milling (conventional) feeds the work opposite to the cutter rotation at the contact zone. This determines the chip thickness variation and therefore the instantaneous cutting force on each tooth.



Given Data / Assumptions:

  • Peripheral milling with steady feed and depth of cut.
  • Negligible runout and uniform tooth spacing.
  • Homogeneous work material.


Concept / Approach:
In up milling, the chip thickness starts nearly zero at entry and increases to a maximum at exit. Cutting force is approximately proportional to instantaneous chip load; thus, force grows from near zero to a peak as the tooth traverses the arc of contact.



Step-by-Step Solution:

Define chip thickness h(θ) in up milling → increases with angular immersion.Force F ∝ h(θ) * width of cut → increases over the tooth path.Therefore, per tooth, force rises from zero at entry to maximum at exit.Contrast with climb milling where chip starts thick and thins to zero.


Verification / Alternative check:
Force dynamometer traces exhibit a rising sawtooth pattern in up milling, matching the increasing chip thickness.



Why Other Options Are Wrong:
“Maximum throughout” and “zero throughout” contradict measured force signals; “decreases from maximum to zero” is the pattern for climb milling, not up milling.



Common Pitfalls:
Mislabeling up vs. down milling when interpreting surface finish and fixture load; always relate force evolution to chip thickness trend.



Final Answer:
Increases from zero to maximum

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