Question
In 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).
Answer :
Correct answer is : 2732
rake angle α = 0∘ , Fc = FT
τs = 500 MPa, shear angle ϕ = 15
width of cut b = 2mm, Fc = ?
Since Fc = FT ⇒ F = N
hence, μ = 1⇒ μ = tanλ
Friction angle: λ = tan-1 (1) = 45
Fc = Fs = ×
Fc = ×
Fc = 2732.05 N ≈ 2732 N
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