GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 37 of 65

Marks: +1, -0

A slot of

25 {mm} × 25 {mm}

is to be milled in a workpiece of

300 {mm}

length using a side and face milling cutter of diameter

100 {mm}

, width

25 {mm}

and having 20 teeth. For a depth of cut

5 {mm}

, feed per tooth

0.1 {mm}

, cutting speed

35 {m} {/} {min}

and approach and over travel distance of

5 {mm}

each, the time required for milling the slot is minutes ______________(round off to one decimal place).

Solution:

Given

{D}=100 {mm}

{L}=300 {mm}

{b}=25 {mm}

{T}=20

{A} \;\text " (Approach) "\;=5 {mm}

{O} \;\text " (Overtravel) "\;=5 {mm}

{d}=5 {mm} .

( {f}) {t}=0 \% {mm}

( {V}) {s}=35 {m} {/} {min}

Since

d<

slot dimension, the complete milling has to be done in 5 passes. Necessary approach

=

Necessary overtravel

=\;{D}/{2}-√{(\;{D}/{2})^{2}-(\;{b}/{2})^{2}}

A N=\;{100}/{2}-√{(\;{100}/{2})^{2}-(\;{25}/{2})^{2}}

{AN}= {A} .587 {mm}

Time Per Cut

\;\; =\;{L+A N+A+O}/{f_{T} N T}

V=∩ D N

N=\;{V}/{π D}=\;{35}/{π × 0.1}

N=111.4 \;\text " RPM "\;

( {T})_{\;\text "percut "\;}=\;{300+1.587+5+5}/{0.1 × 111.4 × 20}

( {T})

per cut

=1.398

min.
Total time

=( {T})_{\;\text "per cut "\;} ×

Number of cut
Total time

=1.398 × 5

=6.99

minutes

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