Question
In an orthogonal machining with a single point cutting tool of rake angle 10°, the uncut chip thickness and the chip thickness are 0.125 mm and 0.22 mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is (round off to two decimal places).
Answer :
0.74
α = 10°
t = 0.125 mm
tc = 0.22 mm
r = t/tc = 0.125/ 0.22 = 0.5682
tan φ = r cosα / (1- r sinα) = 0.5682 cos10° /(1-0.5682sin10°)
φ = 31.83°
Using Merchant’s thoery,
φ = 45° + α/2 - β/2
31.83° = 45° + 10°/2 - β/2
β = 36.34°
μ = tan β = tan 36.34°
= 0.7356 ≈ 0.74
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