CHAPTER 22 - Machining Processes used to Produce Round Shape

EXAMPLE 22.1 - PG NO 600

In [2]:
#example 22.1
#page no. 600 
# Given that
l=6.#in inch Length of rod  
di=1./2.#in inch initial diameter of rod
df=0.480#in inch final diameter of rod
N=400.#in rpm spindle rotation
Vt=8#in inch/minute axial speed of the tool

# Sample Problem on page no. 600

print("\n # Material Removal Rate and Cutting Force in Turning # \n")

V=3.14*di*N
print'%s %d %s' %("\n\n Cutting speed=",V," m/min")

v1=3.14*df*N#cutting speed from machined diameter
d=(di-df)/2#depth of cut
f=Vt/N#feed
Davg=(di+df)/2.
MRR=3.14*Davg*d*f*N 
print'%s %.6f %s' %("\n\n Material Removal Rate =",MRR,"=in^3/min")

t=l/(f*N)
print'%s %.6f %s' %("\n\n Cutting time=",t," min")

P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3
print'%s %.6f %s' %("\n\n Cutting power=",P,"hp")

Fc=((P*396000)/(N*2*3.14))/(Davg/2.)
print'%s %d %s' %("\n\n Cutting force=",Fc,"lb")

 # Material Removal Rate and Cutting Force in Turning # 



 Cutting speed= 628  m/min


 Material Removal Rate = 0.123088 =in^3/min


 Cutting time= 0.750000  min


 Cutting power= 0.180349 hp


 Cutting force= 116 lb

EXAMPLE 22.2 - PG NO 632

In [3]:
#example 22.2
#page no. 632
# Given that  
d=10.#in mm diameter of drill bit
f=0.2#in mm/rev feed
N=800#in rpm spindle rotation

# Sample Problem on page no. 632

print("\n # Material Removal Rate and Torque in Drilling # \n")

MRR=(((3.14*(d**2))/4)*f*N)/60.
print'%s %d %5s' %("\n\n Material Removal Rate ",MRR,"=mm^3/sec")


#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken
T=(MRR*0.5)/((N*2.*3.14)/60.)
print'%s %.6f %s' %("\n\n Torque on the drill ",T,"=Nm")

 # Material Removal Rate and Torque in Drilling # 



 Material Removal Rate  209 =mm^3/sec


 Torque on the drill  1.250000 =Nm