In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principle cutting edge angle is 90⁰, the shear angle in degree is
Answer: B
Given: D = 150 mm, V = 90 m/min, f = 0.24 mm/rev.
d = 2mm, tc = 0.48 mm, α = 0, λ= 90⁰
Uncut chip thickness, t = f sin λ = 0.24 × sin90⁰ = 0.24 mm
Chip thickness ratio, r = t/tc = (0.24/ 0.48) = 1/2
From merchant’s theory,
Shear angle, tanΦ = rcos α /(1 - sin α) = 0.5cos0⁰/ (1 - 0.5sin0⁰) = 0.5
Φ = tan⁻⊃1; (0.5) = 26.56⁰