Question

A 15.24 cm long, 1.27 cm diameter 304 stainless steel rod is being reduced in diameter to 1.2192 cm by turning on a lathe. The spindle rotates at N=400 rpm, and the tool is travelling at an axial speed of 20.32 cm/min. Calculate the cutting speed, material removal rate, power dissipated, and cutting force.

Answer 1

Step-by-step explanation:

Formulas:

v=fNn

`v=fNn\\V=\pi D_0N\\d= ((D_0-D_r))/(2)\\D_(avg)=((D_0+D_r))/(2)\\t(turning)=(I)/(fN)\\t(miling)=((I + I_c))/(v)`

`MRR(turning)= \pi D_(avg)dfN\\MRR(miling) = wdv\\Torque = 0.5(F_cD_(avg))\\Power = (Torque)\omega\\\omega=2\piN`

where
`D_0=1.27cm=12.7mm`;
`N=400rpm`;

The cutting speed is the tangential speed of the workpiece

The maximum cutting speed is at the outer diameter and is obtained from the expression:

`V=\pi D_0N=(\pi (12.7)(400))/(1000)=15.96m/min`

From the above given information, depth of cut is,

`f=(20.32* 10)/(400)=0.508mm/rev`

and the feed is,

`d=((12.7* 10)-(1.2192* 10))/(2)=0.254mm`

According to the given equation, the material removal rate (MRR) is,

`MRR=\pi D_(avg)dfN=\pi (12.446)(0.254)(0.508)(400)=2018.07mm^3/min=2.02* 10^(-6)m^3/min`

Actual time to cut, according to given formular,

`t(turning)=(I)/(fN)=((15.24* 10))/((0.508)(400))=0.75min`

The power required can be calculated by referring to Table 21.2 and taking an average value for stainless steel as 4w-s/mm3. Therefore, the power dissipated is,

`Power=((4)(2018.07))/(60)=134.54W`

Since 1W = 60Nm/min, the power dissipated is 8072.4Mm/min.

The cuting force is the tangential force exerted by the tool. Power is the product of torque T and the rotational speed in radius per unit time; hence,

`T= (8072.4)/((2\pi)(400))=3.21N-m\\\\T_c=((3.21)(1000))/((12.446)/(2))=515.83N`