Linked Data:
In a orthogonal cutting experiment, an HSS tool having the following tool signature in the orthogonal reference system (ORS) has been used : $0 - 10 - 7 - 710 - 75 - 1$
Given: Width of cut $= 3.6 m m$ Shear strength of workpiece material $= 460 N / m m^{2}$ ; Depth of cut $= 0.25 m m$ Coefficient of friction at chip tool interface $= 0.7$

Minimum power requirement (in $k W$ ) at a cutting speed of $150 m / m m$ is

2.5

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3.25

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3.25

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3.45

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Solution

The correct option is B $3.25$
Shear force:
$F_{s} = \frac{τ_{v} \times A_{0}}{s i n ϕ}$
$= \frac{460 \times 3.6 \times 0.25}{s i n {32.5}^{o}} = 770 N$
Cutting force:
$F_{c} = \frac{F_{s}}{c o s (ϕ + β - α)} c o s (β - α)$
$= \frac{770}{c o s {57.5}^{o}} \times c o s 25^{o} = 1299 N$

Power:
$P = F_{C} \times V_{C}$
$= 1299 \times \frac{150}{60} = 3.25 k W$

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