Question

The tool life decreases from 0.8 min to 0.2 min due to the increase in cutting speed from 60 m/min to 120 m/min in a turning operation.
Compute is the value of the cutting speed at 0.4 min.

a-)Determine the spindle speed in rpm.

b-)Find the required power for the unit force.

c-)The friction between tool and workpiece varies 0.25-0.35 during the process.
d-)Calculate the rate of maximum force to minimum force.

Answer 1

Answer:

The value of the cutting speed = 84.85 m/min

a.)The value of the spindle speed = 270.08 rpm

b.)The value of the Power for the unit force = 1.414 W

c.)$u_{max}$ = 0.35,  $u_{min}$ = 0.25

d.)The rate of maximum force to minimum force = 1.4

Explanation:

As given, T₁ = 0.8 min

T₂ = 0.2 min

V₁ = 60 m/min

V₂ = 120 m/min

Now, as we know that Taylor's tool life equation -

VTⁿ = k

where V = speed in m/min

           T = time in min

            n = Index

            k = constant

Now,

V₁T₁ⁿ = V₂T₂ⁿ

⇒60 (0.8)ⁿ = 120(0.2)ⁿ

⇒ (0.8)ⁿ = 2(0.2)ⁿ

⇒ ($\frac{0.8}{0.2}$ )ⁿ = 2

⇒4ⁿ = 2

Taking log both side , we get

⇒n log(4) = log(2)

⇒n = $\frac{log(2)}{log(4)}$ = 0.5

Now,

As given , T = 0.4 min

We know,

VTⁿ = V₁T₁ⁿ

⇒V (0.4)ⁿ = 60(0.8)ⁿ

⇒V = 60($\frac{0.8}{0.4}$ )ⁿ

⇒V = 60(2)ⁿ

⇒V = 60($2^{0.5}$)

⇒V = 84.85 m/min

∴ The value of the cutting speed = 84.85 m/min

a.)

As we know ,

V = π×d×N

Let d = 100 mm = 0.1 m

⇒ 84.85 = π×0.1×N

⇒ N = $\frac{84.85}{0.1\pi } = 270.08$ rpm

∴ The value of the spindle speed = 270.08 rpm

b.)

As we know,

P = F×V

⇒P = 1×$\frac{84.85}{60} = 1.414$ W per unit force

∴ The value of the Power for the unit force = 1.414 W

c.)

As given - The friction between tool and workpiece varies 0.25-0.35 during the process.

⇒$u_{max}$ = 0.35

  $u_{min}$ = 0.25

d.)

Rate of maximum force , $F_{max}$ = $u_{max}$ × Normal force

Rate of minimum force , $F_{min}$ = $u_{min}$ × Normal force  

∴ we get

$\frac{F_{max} }{F_{min} } = \frac{u_{max} }{u_{min} } = \frac{0.35}{0.25}$ = 1.4

∴ we get

The rate of maximum force to minimum force = 1.4