Linked Data:
A $ϕ 40 m m$ job is subjected to orthogonal turning by $a + 10^{o}$ 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 $25^{o}$

If the friction angle at the tool chip interface is $58^{o} 10^{'}$ and the cutting force components measured by a dynamometer are $600 N$ and $200 N$ the power loss due to friction (in $k N m / m i n$ ) is approximately

20

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18

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16

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350

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Solution

The correct option is D $350$
Chip thickness ratio:
$r = \frac{s i n ϕ}{c o s (ϕ - α)}$
$= \frac{s i n 25^{o}}{c o s 15^{o}} = 0.437$

$r = \frac{V_{f}}{V_{c}}$
$\Rightarrow V_{f} = r V_{c} = 0.437 \times 62.8$
$= 27.5 m / m i n$

$β = 58^{o} 10$
$F_{c} = 600 N, F_{T} = 200 N$
$F = \frac{F_{c}}{c o s (β - α)} \times s i n β$
$= \frac{600}{c o s 48^{o} 10^{'}} \times s i n 58^{o} 10^{'}$

$= 764 N$

Loss of energy due to friction
$= F \times V_{f}$
$= \frac{764 \times 27.5}{60} = 350.3 W$

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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