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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 90°, the shear angle in degree is

Answer & Solution
Correct Answer: Option B
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Dhiraj Ghorpade
Dhiraj Ghorpade 6 years ago
Option B is correct.

Ans: Shear angle is given as tan@= rcos@ ÷ 1-rsin@@

Where, r= t÷tc,. Tc=0.48
t , fsin(90)= 0.24×1= 0.24


Hence ,. r= 0.24÷0.48= 0.5

given as tan@= rcos@ ÷ 1-rsin@
Put value r , we get @=26.56