Project ForgeA cylindrical bar of 200 mm diameter is turned with a tool of geometry 0°-9°-7°-8°-15°-30°-0.05 inch (ASA system), resulting in cutting force Fc1. If the tool geometry is changed to 0°-9°-7°-8°-15°-0°-0.05 inch (only the side cutting edge angle changes, from 30° to 0°) and all other parameters remain unchanged, the cutting force becomes Fc2. Specific cutting energy (J/mm³) is Uc=U0(t1)^−0.4, where t1 is the uncut chip thickness. The percentage change in cutting force, [(Fc2−Fc1)/Fc1]×100, is __________ (round off to one decimal place).
Explanation
With approach angle ψr=90°−SCEA: t1=f·sinψr, width of cut w=d/sinψr (both feed f and depth d unchanged). Force Fc=Uc×(t1·w)=U0·t1^0.6·w=U0·f^0.6·d·sinψr^(−0.4). At SCEA=30° (ψr=60°): sin60°^(−0.4)≈1.0592. At SCEA=0° (ψr=90°): sin90°^(−0.4)=1. %change=(1−1.0592)/1.0592×100≈−5.59%, i.e. within GATE's key range of −5.8 to −5.5.
Source: GATE ME 2020 Session 2 (ME-2) (2020)
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