6. A 200-mm-long, 75-mm-diameter titanium-alloy rod is being reduced in diameter to 65 mm by turning on a lathe. The spindle rotates at 400 rpm, and the tool is travelingat an axial velocity of 200 mm/min. Calculate the cutting speed, material removal rate, time of cut, power required, and cutting force.

Respuesta :

Answer:

(a) 94.25 m/min

(b) [tex]219911.5 mm^{3}/min[/tex]

(c) 1 min

(d)  10.996 kW and 15.027 kW

(e) 7 kN and 9.57 kN

Explanation:

The maximum cutting speed, [tex]v_c[/tex] is observed at the outer diameter hence

[tex]v_c=N\piD_o[/tex] where N is the rate of rotation and [tex]D_o[/tex] is the outer diameter. Substituting 75mm for [tex]D_o[/tex] and 400 rpm for N we obtain

[tex]v_c=\frac {400\times \pi\times 75}{1000}=94.24778 m/min\approx 94.25 m/min[/tex]

(b)

The material removal rate, MRR is given by

[tex]MMR=\pi\times D_{avg}\times d\times f \times N[/tex] where [tex]D_{avg}[/tex] is the average between the inner and outer diameters, d is the depth of cut, f is the feed which is given by [tex]\frac {V}{N}[/tex] where V is the axial velocity

In this case, the average diameter is [tex]\frac {75+65}{2}=70mm[/tex]

The feed f is [tex]f=\frac {200}{400}=0.5[/tex]  

The depth of cut is [tex]d=\frac {75-65}{2}=5mm[/tex]

Therefore, [tex]MRR=\pi\times 70mm\times 5\times 0.5 \times 400 \approx 219911.5 mm^{3}/min[/tex]

(c)

Time of cut is given by

[tex]T=\frac {L}{fN}[/tex] where L is the length of rod. Substituting L for 200mm, f as seen in part b is 0.5 and 400 for N we obtain

[tex]T=\frac {200}{0.5*400}=1 min[/tex]

(d)

The power is obtain by multiplying specific energy by material removal rate. Assuming specific energy range of [tex]3 ws/mm^{3}[/tex] to [tex]4.1 ws/mm^{3}[/tex]  then

Power, [tex]P1=\frac {3\times 219911.5}{60}=10995.57 W\approx 10.996 kW[/tex]

Power, [tex]P2=\frac {4.1\times 219911.5}{60}=15027.28 W\approx 15.027 kW[/tex]

Therefore, power ranges between 10.996 kW and 15.027 kW

(e)

Cutting force is given by

[tex]F_c=\frac {P}{v_c}[/tex] where P is power and [tex]v_c[/tex] is already calculated in part a

First, [tex]v_c[/tex] is converted to m/s hence [tex]v_c=\frac {400\times \pi\times 75}{1000\times 60}= 1.570796\approx 1.57 m/s[/tex]

From the power range in part d,

[tex]F_{c1}=\frac {10.996 kW}{1.57}= 7.003822\approx 7 kN[/tex]

[tex]F_{c2}=\frac {15.027 kW}{1.57}= 9.571338\approx 9.57 kN[/tex]

Therefore, the cutting force ranges from 9.57 kN and 7 kN