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

Answer:

(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

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

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

(b)

The material removal rate, MRR is given by

$MMR=\pi\times D_{avg}\times d\times f \times N$ 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, $MRR=\pi\times 70mm\times 5\times 0.5 \times 400 \approx 219911.5 mm^{3}/min$

(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, $P1=\frac {3\times 219911.5}{60}=10995.57 W\approx 10.996 kW$

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

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 $v_c=\frac {400\times \pi\times 75}{1000\times 60}= 1.570796\approx 1.57 m/s$

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