CHAPTER 32 - Tribology Friction Wear and Lubrication

EXAMPLE 32.1 - PG NO 886

In [2]:
#example 32.1
#page no. 886
import math
# Given that
hi=10.#in mm height of specimen
ODi=30.#in mm outside diameter 
IDi=15.#in mm inside diameter 
ODf=38.#in mm outside diameter after deformaton
#Specimen is reduced in thickness by 50%
hf=(50./100.)*hi

# Sample Problem on page no. 886

print("\n # Determination of Cofficient of Friction # \n")

IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf
ID=((IDi-IDf)/IDi)*100#change in internal diameter 

print'%s %d %s %s' %("\n\n With a 50 percent reduction in height and a ",ID,"%"," reduction in internal diameter, from the book data Cofficient of Friction = 0.21") 
 # Determination of Cofficient of Friction # 



 With a 50 percent reduction in height and a  35 %  reduction in internal diameter, from the book data Cofficient of Friction = 0.21