GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 45 of 65

Marks: +1, -0

An orthogonal cutting operations is being carried out in which uncut thickness is

0.010

{mm}

, cutting speed is

130 {m} {/} {min}

, rake angle is

15^{∘}

and width of cut is

6 {mm}

. It is observed that the chip thickness is

0.015 {mm}

, the cutting force is

60 {N}

and the thrust force is

25 {N}

. The ratio of friction energy to total energy is (correct to two decimal places)

Solution:

t=0.010 {mm}

v=130 {m} {/} {min}

α=15^{∘}

b=6 {mm}

t_{c}=0.015 {mm}

F_{c}=60 {N}

F_{t}=25 {N}

F=F_{c} \sin α+F_{t} \cos α

=60 \sin 15+25 \cos 15=39.6773 {N}

Where F represents frictional force
Ratio of frictional energy to total energy

=\;{F}/{F_{c}}, \;{V_{c}}/{V}=\;{F}/{F_{c}}(\;{t}/{t_{c}})

[ ∵ \;{t}/{t_{c}}=\;{V_{c}}/{V}=r]

=\;{39.6773}/{60} × \;{0.010}/{0.015}=0.4408

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