Chapter 15: Fundamentals of Metalworking¶

Example 15.1, Mechanics of Metal Working, Page No. 506¶

In [9]:
from math import log

#For Bar which is double in length
#variable declaration 1
L2=2;
L1=1;

#calculation 1
e=(L2-L1)/L1;
e1=log(L2/L1);
r=1-L1/L2;

#result 1
print('\nEnginering Strain = %g\nTrue Strain = %g\nReduction = %g')%(e,e1,r);

#For bar which is halved in length
#variable declaration 2
L1=1;
L2=0.5;

#calculation 2
e=(L2-L1)/L1;
e1=log(L2/L1);
r=1-L1/L2;

#result 2
print('\n\nEnginering Strain = %g\nTrue Strain = %g\nReduction = %g')%(e,e1,r);

Enginering Strain = 1
True Strain = 0.693147
Reduction = 1

Enginering Strain = -0.5
True Strain = -0.693147
Reduction = -1


Example 15.2, Mechanics of Metal Working, Page No. 511¶

In [10]:
from scipy.integrate import quad
from math import log

#variable declaration
D0=25.0;
D1=20.0;
D2=15.0;
def integrand(e):
return 200000*e**0.5

#calculation
ep1=log((D0/D1)**2);
ep2=log((D1/D2)**2);

#result
print('\nPlastic work done in 1st step = %g lb/in^2\nPlastic work done in 2nd step = %g lb/in^2\n')%(U1,U2);

Plastic work done in 1st step = 39752.1 lb/in^2
Plastic work done in 2nd step = 97934.8 lb/in^2



Example 15.3, Hodography, Page No. 517¶

In [11]:
from math import sin
from math import radians

#variable declaration
alpha=60;

#calculation
mu=1/sin(r);
p_2k=mu*5/2;

#result
print('Pressure  = %g')%(p_2k);

Pressure  = 2.88675


Example 15.4, Temperature in Metalworking, Page No. 526¶

In [12]:
#variable declaration
Al_s=200;
Al_e=1;
Al_p=2.69;
Al_c=0.215;
Ti_s=400;
Ti_e=1;
Ti_p=4.5;
Ti_c=0.124;
J=4.186;
b=0.95;

#calculation
Al_Td=Al_s*Al_e*b/(Al_p*Al_c*J);
Ti_Td=Ti_s*Ti_e*b/(Ti_p*Ti_c*J);

#result
print('\nTemperature Rise for aluminium = %g C\nTemperature Rise for titanium = %g C\n')%(Al_Td,Ti_Td);

Temperature Rise for aluminium = 78.4808 C
Temperature Rise for titanium = 162.686 C



Example 15.5, Friction and Lubrication, Page No. 546¶

In [13]:
from math import sqrt

#variable declaration
Do=60;
Di=30;
def1=70;
def2=81.4;
h=10;
a=30;

#calculation1
di=sqrt((Do**2-Di**2)*2-def1**2);
pr=(Di-di)/Di*100;
m=0.27;
p_s=1+2*m*a/(sqrt(3)*h);

#result 1
print('\nFor OD after deformation being 70 mm, Di = %g mm\nPrecent change in inside diameter = %g percent\nPeak pressure = %g')%(di,pr,p_s);

#calculation 2
di=sqrt(def2**2-(Do**2-Di**2)*2);
pr=(Di-di)/Di*100;
m=0.05;
p_s=1+2*m*a/(sqrt(3)*h);

#result 2
print('\n\n\n\nFor OD after deformation being 81.4 mm, Di = %g mm\nPrecent change in inside diameter = %g percent\nPeak pressure = %g')%(di,pr,p_s);

For OD after deformation being 70 mm, Di = 22.3607 mm
Precent change in inside diameter = 25.4644 percent
Peak pressure = 1.93531

For OD after deformation being 81.4 mm, Di = 35.0137 mm
Precent change in inside diameter = -16.7124 percent
Peak pressure = 1.17321