GATE 2020 ME (ME2) – Question 53
A 200 mm diameter cylindrical bar is turned using ASA tool geometry 0°–9°–7°–8°–15°–30°–0.05 inch, giving cutting force $F_{c1}$. Geometry changes to 0°–9°–7°–8°–15°–0°–0.05 inch; all other parameters stay fixed. Specific cutting energy is $U_c=U_0t_1^{-0.4}$ J/mm³, where $t_1$ is uncut thickness in mm and $U_0$ is constant. Find $100(F_{c2}-F_{c1})/F_{c1}$, to one decimal.
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Correct answer: -5.8 to -5.5
Explanation
Uncut chip thickness is t1=f cosψ. Changing ψ from30° to0° multiplies thickness by1/cos30°. Uncut area remains constant, so force ratio equals specific-energy ratio $(1/\cos30^\circ)^{-0.4}=(\cos30^\circ)^{0.4}=0.94409$. Percentage change is100(0.94409−1)=**−5.6%**.