GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 47 of 65

Marks: +1, -0

Bars of

250 {mm}

length and

25 {mm}

diameter are to be turned on a lathe with a feed of

0.2

{mm} {/} {rev}

. Each regrinding of the tool costs Rs.

20 .

The time required for each tool change is 1 min. Tool life equation is given as

{VT}^{0.2}=24

(where cutting speed

{V}

is in

{m} {/} {min}

and tool life

T

is in min). The optimum tool cost per piece for maximum production rate is Rs. _______________ (round off to 2 decimal places).

Solution:

Turning on Lath :
Length

( {L})=250 {mm}

Diameter (D)

=25 {mm}

Feed

(f)=0.2 {mm}{/} {rev}

Regrinding tool cost

= Rs 20

/ regrind Tool change time

(T_{c})=1 {min}

.
Tool life equation

V T^{0.2}=24

Tool life for optimum production rate:
since

{T}= {T}_{ {c}}(\;{1}/{ {n}}-1)

=1(\;{1}/{0.2}-1)

Tool life

(T)=4

min
Now

{VT}^{0.2}=24

{V}=18.188 {m} {/} {min}

Now

V T^{0.2}=24

{V}=18.188 {m}{/} {min}

Now

V=\;{π D N}/{1000}

18.188=\;{π × 25 × N}/{1000}

{N}=23.158 {rpm}

Thus, Machining time

( {t}_{ {m}})=\;{ {L}}/{ {fN}}=\;{250 {mm}}/{0.2 {mm} {/} {rev} × 231.58}

{tm}=5.3975 {min}

Thus, optimum tool cost

(C_{T})=\;{5.3975}/{4} × 20

=26.9

Rs.

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