# CHAPTER 25 - Abrasive Machining and Finishing Operations¶

## EXAMPLE 25.1 - PG NO 713¶

In :
#example 25.1
#page no. 713
import math
# Given that
D=200#in mm Grinding Wheel diameter
d=0.05#in mm depth of cut
v=30#m/min workpiece velocity
V=1800#in m/min wheel velocity

# Sample Problem on page no. 713

print("\n # Chip Dimensions in Surface Grinding # \n")

l=math.sqrt(D*d)
l1=l/2.54*(10**-1)
print'%s %.6f %s'%("\n\n Undeformed Chip Length =",l1,"mm")

#the answer in the book is approximated to 0.13 in

#assume
C=2.#in mm
r=15.
t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))
t1=t/2.54*(10**-1)
print'%s %.6f %s' %("\n\n Undeformed chip Thickness =",t1,"in")

#the answer in the book is approximated to 0.00023in

 # Chip Dimensions in Surface Grinding #

Undeformed Chip Length = 0.124499 mm

Undeformed chip Thickness = 0.000233 in


## Example 25.2 - PG NO 715¶

In :
#example 25.2
#page no. 715
# Given that
D=10.#in inch Grinding Wheel diameter
N=4000.#in rpm
w=1.#in inch
d=0.002#in inch depth of cut
v=60.#inch/min feed rate of the workpiece

# Sample Problem on page no. 715

print("\n # force in Surface Grinding # \n")

Mrr=d*w*v#material removal rate
#for low carbon steel , the specific energy is 15hp min/in3
u=15.#in hp min/in3
P=u*Mrr*396000.#in lb/min
Fc = P/(2*3.14*N*(D/2.))

print'%s %.6f %s' %("\n\n Cutting Force =",Fc,"lb")

Fn = Fc+(30./100.)*Fc

print'%s %.6f %s' %("\n\n Thrust Force =",Fn,"lb")

 # force in Surface Grinding #

Cutting Force = 5.675159 lb

Thrust Force = 7.377707 lb