Project ForgeIn an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).
Explanation
Since Fc=Ft, Merchant's circle gives cos(β−α)=sin(β−α), so β−α=45°; with α=0°, friction angle β=45°. Shear area As=t1w/sinφ=0.5×2/sin15°≈3.8637 mm². Shear force Fs=τs·As=500×3.8637≈1931.85 N. Merchant's circle: Fc=Fs·cos(β−α)/cos(φ+β−α)=1931.85×cos45°/cos(15+45)=1931.85×0.7071/0.5≈2732 N.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
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