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Force Analysis of Orthogonal Machining Part-1

Common Data:I...

Question

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 orthogonal rake angle of the cutting toll in degree is

zero

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Solution

The correct option is A zero
Diameter of bar, $d_{i} = 100 m m$
Feed, $f = 0.25 m m / r e v$
Depth of cut, $d = 4 m m$
$C = 90 m / m i n$
Cutting force $= 1500 N$
Since cutting force, is perpendicular to the friction force,
By Merchant's circle

This is the case of, when $α = 0$

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Similar questions

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

Q. In orthogonal turning of a low carbon steel bar of diameter

150 m m

uncoated carbide tool the cutting velocity is

90 m / m i n

. The feed is

0.24 m m / r e v

and the depth of cut is

2 m m

. The chip thickness obtained is

0.48 m m

. If the orthogonal rake angle is zero and the principal cutting edge angle is

90^{o}

, the shear angle in degree is

Q. The following data was observed during orthogonal cutting a steel work piece of 60 mm diameter.
Cutting speed = 90 m/min,
Feed= 0.15 mm/rev,
Depth of cut 5 mm,
Length of chip /rev = 80.8 mm,
Rake angle = 10

^{\circ}

,
Clearance angle = 8

^{\circ}

Tangential force = 2200 N.
Thrust force 1200 N

The energy lost in friction 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

Q. In orthogonal turning of low carbon steel pipe with principal cutting edge angle of

90^{o}

, the main cutting force is

1000 N

and the feed force is

800 N

. The shear angle is

25^{o}

and the orthogonal rake angle is zero. Employing Merchant's theory, the ratio of friction force to normal force acting on the cutting tool is

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The correct option is A zeroDiameter of bar, di=100mm Feed, f=0.25mm/rev Depth of cut, d=4mm C=90m/min Cutting force =1500N Since cutting force, is perpendicular to the friction force, By Merchant's circle This is the case of, when α=0

The correct option is A zero
Diameter of bar, $d_{i} = 100 m m$
Feed, $f = 0.25 m m / r e v$
Depth of cut, $d = 4 m m$
$C = 90 m / m i n$
Cutting force $= 1500 N$
Since cutting force, is perpendicular to the friction force,
By Merchant's circle

This is the case of, when $α = 0$