The rake angle of a cutting tool is $15^{o}$ , shear angle $45^{o}$ and cutting velocity 35 m/min. What is the velocity of the chip along the tool face?

28.5 m / m i n

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27.3 m / m i n

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25.3 m / m i n

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23.5 m / m i n

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Solution

The correct option is A $28.5 m / m i n$
Given:
$α = 15^{\circ}, ϕ = 45^{\circ}, V_{C} = 35 m / m i n$

Chip thickness ratio:
$r = \frac{s i n ϕ}{c o s (ϕ - α)} = \frac{s i n 45^{o}}{c o s 30^{o}}$
$= 0.816$

The ratio of chip flow velocity and cutting velocity is equal to chip thickness ratio:
$\frac{V_{f}}{V_{C}} = r$
$V_{f} = r V_{C} = 0.816 \times 35$
$= 28.58 m / m i n$
Velocity of the chip at the tool face:
$V_{f} = 28.58$ m\min

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The rake angle of a cutting tool is 15o, shear angle 45o and cutting velocity 35 m/min. What is the velocity of the chip along the tool face?

The rake angle of a cutting tool is $15^{o}$ , shear angle $45^{o}$ and cutting velocity 35 m/min. What is the velocity of the chip along the tool face?