In an orthogonal machining operation,
if U = specific cutting energy, β= friction angle, ϕ= shear plane angle, α= Tool rake angle, then the shear stress (τ) will be expressed as:
A
Usinϕcos(β−α)cos(ϕ+β−α)
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B
Ucosαcos(ϕ+β−α)cos(β−α)
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C
Usinϕcos(ϕ+β−α)cos(β−α)
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D
Usinαcos(ϕ−α)cos(β−α)
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Solution
The correct option is CUsinϕcos(ϕ+β−α)cos(β−α) Shear stress, τ=Fsbt1sinϕ