CHAPTER 14 - Forging of Metals

EXAMPLE 14.1 - PG NO 344

In [1]:
#example 14.1
#page no. 344
# Given that
import math
d=150.#in mm Diameter of the solid cylinder 
Hi=100. #in mm Height of the cylinder
u=0.2 # Cofficient of friction

# Sample Problem on page no. 344

print("\n # Calculation of forging force # \n")

#cylinder is reduced in height by 50%
Hf=100./2.
#Volume before deformation= Volume after deformation
r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder
E=math.log(Hi/Hf)#absolute value of true strain
#given that cylinder is made of 304 stainless steel
Yf=1000. #in Mpa flow stress of the material from data in the book
F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force
F1=F/(10.**6.)
print'%s %d %s' %("\n\n Forging force = ",F1,"MN")
 # Calculation of forging force # 



 Forging force =  45 MN