Question

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.

Answer 1

(a) 94.25 m/min

(b)
`219911.5 mm^(3)/min`

(c) 1 min

(d) 10.996 kW and 15.027 kW

(e) 7 kN and 9.57 kN

Step-by-step explanation:

The maximum cutting speed,
`v_c` is observed at the outer diameter hence

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

`v_c=\frac {400* \pi* 75}{1000}=94.24778 m/min\approx 94.25 m/min`

(b)

The material removal rate, MRR is given by

`MMR=\pi* D_(avg)* d* f * N` where
`D_(avg)` is the average between the inner and outer diameters, d is the depth of cut, f is the feed which is given by
`\frac {V}{N}` where V is the axial velocity

In this case, the average diameter is
`\frac {75+65}{2}=70mm`

The feed f is
`f=\frac {200}{400}=0.5`

The depth of cut is
`d=\frac {75-65}{2}=5mm`

Therefore,
`MRR=\pi* 70mm* 5* 0.5 * 400 \approx 219911.5 mm^(3)/min`

(c)

Time of cut is given by

`T=\frac {L}{fN}` 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

`T=\frac {200}{0.5*400}=1 min`

(d)

The power is obtain by multiplying specific energy by material removal rate. Assuming specific energy range of
`3 ws/mm^(3)` to
`4.1 ws/mm^(3)` then

Power,
`P1=\frac {3* 219911.5}{60}=10995.57 W\approx 10.996 kW`

Power,
`P2=\frac {4.1* 219911.5}{60}=15027.28 W\approx 15.027 kW`

Therefore, power ranges between 10.996 kW and 15.027 kW

(e)

Cutting force is given by

`F_c=\frac {P}{v_c}` where P is power and
`v_c` is already calculated in part a

First,
`v_c` is converted to m/s hence
`v_c=\frac {400* \pi* 75}{1000* 60}= 1.570796\approx 1.57 m/s`

From the power range in part d,

`F_(c1)=\frac {10.996 kW}{1.57}= 7.003822\approx 7 kN`

`F_(c2)=\frac {15.027 kW}{1.57}= 9.571338\approx 9.57 kN`

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