You visited us 10 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Production

Tool Life - II

In a cutting ...

Question

In a cutting test with 0.3 mm flank wear as tool failure criterion, a tool life of 10 min was obtained at a cutting velocity of 20 m/min. Taking tool life exponent as $0.25$ , the tool life in minutes at 40 m/min of cutting velocity will be

0.506

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.625

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.140

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.002

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.625$
$n = 0.25, T_{1} = 10 m i n, V_{1} = 20 m / m i n$
If $V_{2} = 40 m / m i n$
Taylor's tool life equation:
$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$
$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{1 / n} = 10 {(\frac{20}{40})}^{1 / 0.25}$
$= 10 {(\frac{1}{2})}^{4} = 0.625 m i n$

Suggest Corrections

Similar questions

Q. Tool life of a

10

hours is obtained when cutting with a single point tool at

63 m / m i n

. If Taylor's constant

C = 257.35,

tool life on doubling the velocity will be

Q. In an orthogonal machining operation, the tool life obtained is

10 m i n

at a cutting speed of

100 m / m i n

while at

75 m / m i n

cutting speed, the tool life is

30 m i n

. The value of index

(n)

in the Taylor's tool life equation is

Q. During a tool-wear test with a HSS tool, the following data are obtained:

Tool life	Cutting speed
30 min	20 m/min
5 min	75 m/min

The tool life of the tool if cutting velocity is 50 m/min, is

Q. Linked Question:
In a machining experiment, tool life was found to vary with the cutting speed in the following manner:

Cutting speed $(m / m i n)$	Tool life $(m i n u t e s)$
$60$	$81$
$90$	$36$

The exponent

(n)

and constant

(k)

of the Taylor's tool life equation are

Q. During plain turning of medium carbon steel by using HSS tool, a cutting velocity of 25 m/min was observed for 1min tool life and 20 m/min for 5min tool life. The Taylor's constant for the HSS Tool is

Join BYJU'S Learning Program

In a cutting test with 0.3 mm flank wear as tool failure criterion, a tool life of 10 min was obtained at a cutting velocity of 20 m/min. Taking tool life exponent as $0.25$ , the tool life in minutes at 40 m/min of cutting velocity will be

The correct option is B $0.625$
$n = 0.25, T_{1} = 10 m i n, V_{1} = 20 m / m i n$
If $V_{2} = 40 m / m i n$
Taylor's tool life equation:
$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$
$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{1 / n} = 10 {(\frac{20}{40})}^{1 / 0.25}$
$= 10 {(\frac{1}{2})}^{4} = 0.625 m i n$

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US

In a cutting test with 0.3 mm flank wear as tool failure criterion, a tool life of 10 min was obtained at a cutting velocity of 20 m/min. Taking tool life exponent as 0.25, the tool life in minutes at 40 m/min of cutting velocity will be

The correct option is B 0.625n=0.25,T1=10min,V1=20m/min If V2=40m/min Taylor's tool life equation: V1Tn1=V2Tn2 T2=T1(V1V2)1/n=10(2040)1/0.25 =10(12)4=0.625min

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US

In a cutting test with 0.3 mm flank wear as tool failure criterion, a tool life of 10 min was obtained at a cutting velocity of 20 m/min. Taking tool life exponent as $0.25$ , the tool life in minutes at 40 m/min of cutting velocity will be

The correct option is B $0.625$
$n = 0.25, T_{1} = 10 m i n, V_{1} = 20 m / m i n$
If $V_{2} = 40 m / m i n$
Taylor's tool life equation:
$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$
$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{1 / n} = 10 {(\frac{20}{40})}^{1 / 0.25}$
$= 10 {(\frac{1}{2})}^{4} = 0.625 m i n$