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

32

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63

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22

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27.5

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Solution

The correct option is D $27.5$
$D = 40 m m, α = 10, N = 500 r p m, ϕ = 25^{o}$

Cutting velocity:
$V = \frac{π D N}{1000} = \frac{π \times 40 \times 500}{1000}$
$= 62.8 m / m i n$

$\frac{V_{c}}{V} = \frac{s i n ϕ}{c o s (ϕ - α)}$

Velocity of the chip:
$V_{c} = 27.5 m / m i n$

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Linked Data: A ϕ40mm job is subjected to orthogonal turning by +10o rake angle tool at 500rev/min. By direct measurement during the cutting operation, the shear angle was found equal to 25o. The velocity (in m/min) with which the chip flows on the tool face is

The correct option is D 27.5D=40mm,α=10,N=500rpm,ϕ=25o Cutting velocity: V=πDN1000=π×40×5001000 =62.8m/min VcV=sinϕcos(ϕ−α) Velocity of the chip: Vc=27.5m/min

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

The correct option is D $27.5$
$D = 40 m m, α = 10, N = 500 r p m, ϕ = 25^{o}$

Cutting velocity:
$V = \frac{π D N}{1000} = \frac{π \times 40 \times 500}{1000}$
$= 62.8 m / m i n$

$\frac{V_{c}}{V} = \frac{s i n ϕ}{c o s (ϕ - α)}$

Velocity of the chip:
$V_{c} = 27.5 m / m i n$