If $α$ is the rake angle of the cutting tool, $ϕ$ is, the shear angle, V is the cutting velocity and $V_{c}$ is the velocity of chip relative to tool, the velocity of chip sliding along the shear plane is given by

\frac{V sin ϕ}{cos (ϕ - α)}

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\frac{V cos α}{sin (ϕ - α)}

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\frac{V_{C} cos α}{sin (ϕ)}

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\frac{V_{C} cos α}{cos (ϕ - α)}

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Solution

The correct option is C $\frac{V_{C} cos α}{sin (ϕ)}$
Velocities in metal cutting, by applying sine rule,
$\frac{V}{cos (ϕ - α)} = \frac{V_{c}}{sin ϕ} = \frac{V_{s}}{cos α}$

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Similar questions

Q. If a is the rake angle of a cutting tool,

ϕ

is the shear angle, then the ratio of velocity of chip to the shear velocity is given by

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

Linked Data:
A $ϕ 40 m m$ job is subjected to orthogonal turning by $+ 10^{o}$ rake angle tool at $500 r e v / m i n$ . By direct measurement during the cutting operation, the shear angle was found equal to $25^{o}$ .

The velocity (in $m / m i n$ ) with which the chip flows on the tool face is

Q. Common Data:
An orthogonal operation is carried out at

20 m / m i n

cutting speed, using a cutting tool of rake angle

15

degree. The chip thickness is

0.4 m m

and uncut chip thickness is

0.2 m m

The shear angle (in degrees) is

Common Data:
An orthogonal operation is carried out at $20 m / m i n$ cutting speed, using a cutting tool of rake angle $15$ degree. The chip thickness is $0.4 m m$ and uncut chip thickness is $0.2 m m$

The chip velocity (in $m / m i n$ ) is

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If α is the rake angle of the cutting tool, ϕ is, the shear angle, V is the cutting velocity and Vc is the velocity of chip relative to tool, the velocity of chip sliding along the shear plane is given by

The correct option is C VCcosαsin(ϕ)Velocities in metal cutting, by applying sine rule, Vcos(ϕ−α)=Vcsinϕ=Vscosα

If $α$ is the rake angle of the cutting tool, $ϕ$ is, the shear angle, V is the cutting velocity and $V_{c}$ is the velocity of chip relative to tool, the velocity of chip sliding along the shear plane is given by

The correct option is C $\frac{V_{C} cos α}{sin (ϕ)}$
Velocities in metal cutting, by applying sine rule,
$\frac{V}{cos (ϕ - α)} = \frac{V_{c}}{sin ϕ} = \frac{V_{s}}{cos α}$