Answer:
Explanation:
Formulas:
v=fNn
[tex]v=fNn\\V=\pi D_0N\\d= \frac{(D_0-D_r)}{2}\\D_{avg}=\frac{(D_0+D_r)}{2}\\t(turning)=\frac{I}{fN}\\t(miling)=\frac{(I + I_c)}{v}[/tex]
[tex]MRR(turning)= \pi D_{avg}dfN\\MRR(miling) = wdv\\Torque = 0.5(F_cD_{avg})\\Power = (Torque)\omega\\\omega=2\piN[/tex]
where [tex]D_0=1.27cm=12.7mm[/tex]; [tex]N=400rpm[/tex];
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:
[tex]V=\pi D_0N=\frac{\pi (12.7)(400)}{1000}=15.96m/min[/tex]
From the above given information, depth of cut is,
[tex]f=\frac{20.32\times 10}{400}=0.508mm/rev[/tex]
and the feed is,
[tex]d=\frac{(12.7\times 10)-(1.2192\times 10)}{2}=0.254mm[/tex]
According to the given equation, the material removal rate (MRR) is,
[tex]MRR=\pi D_{avg}dfN=\pi (12.446)(0.254)(0.508)(400)=2018.07mm^3/min=2.02\times 10^{-6}m^3/min[/tex]
Actual time to cut, according to given formular,
[tex]t(turning)=\frac{I}{fN}=\frac{(15.24\times 10)}{(0.508)(400)}=0.75min[/tex]
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,
[tex]Power=\frac{(4)(2018.07)}{60}=134.54W[/tex]
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,
[tex]T= \frac{8072.4}{(2\pi)(400)}=3.21N-m\\\\T_c=\frac{(3.21)(1000)}{\frac{12.446}{2}}=515.83N[/tex]
