GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 16 of 65

Marks: +1, -0

The length, width and thickness of a steel sample are

400 {mm}, 40 {mm}

and

20 {mm}

, respectively. Its thickness needs to be uniformly reduced by

2 {mm}

in a single pass by using horizontal slab milling. The milling cutter (diameter:

100 {mm}

, width:

50 {mm}

) has 20 teeth and rotates at 1200 rpm. The feed per tooth is

0.05 {mm}

. The feed direction is along the length of the sample. If the over-travel distance is the same as the approach distance, the approach distance and time taken to complete the required machining task are

$21 {mm}, 28.9 {s}$ .
$14 {mm}, 21.4 {s}$
$21 {mm}, 39.4 {s}$ .
$14 {mm}, 18.4 {s}$

Validate

Solution:

Depth of cut (d)

= 2

mm
Milling cutter diameter (D)

=100 {mm}

over-travel distance is the same as the approach distance

=√{d(D-d)}=√{2(100-2)}=14 {mm}

Total distance travelled by tool = length of sample + approach distance + overtravel

= 400 + 14 + 14 = 428

mm
Feed per tooth = 0.05 mm
No. of teeth

= 20

So, feed (f)

= 20 × 0.05 = 1

mm/rev
Rotation speed (N)

= 1200rpm

Therefore, machining time

=\;{\;\text " total length "\;}/{\;\text " feed "\; × N}=\;{428}/{1 ×(\;{1200}/{60})}=21.4 {sec}

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