Linked Data:
In a orthogonal cutting experiment, an HSS tool having the following tool signature in the orthogonal reference system (ORS) has been used : $0 - 10 - 7 - 7 - 10 - 75 - 1$
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 $= 0.7$

Shear plane angle (in degrees) for minimum cutting force is

20.5

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24.5

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28.6

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32.5

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Solution

The correct option is D $32.5$
Given:
$w = b = 3.6 m i n, τ_{u} = 460$
$d = t_{1} = 0.2; μ = 0.7$
$α = 10 =$ orthogonal rake angle (ORS)
Shear plane angle for minimum cutting force
$2 ϕ + β - α = 90^{o}$
$ϕ = \frac{90^{o} + α - β}{2}$
$= \frac{90^{o} + 10^{o} - 35^{o}}{2} = {32.30}^{o}$

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Similar questions

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

Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is

Common Data:
In orthogonal turning of a bar of $100 m m$ diameter with a feed of $0.25 m m / r e v$ , depth of cut of $4 m m$ and cutting velocity of $90 m / m i n$ , it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is $1500 N$ .

The normal force acting the chip-tool interface in N is