# Chapter 3: Elements of the Theory of Plasticity¶

### Example 3.1, True Stress and True Strain, Page No. 76¶

In [4]:
from math import pi
from math import log
from math import exp

#variable declaration
D_i=0.505;
L=2;
P_max=20000;
P_f=16000;
D_f=0.425;

#calculation
E_St= P_max*4/(pi*D_i**2);
T_fr_St= P_f*4/(pi*D_f**2);
e_f=log(D_i**2/D_f**2);
e=exp(e_f)-1;

#result
print('\nEngineering Stress at maximum load = %g psi\nTrue Fracture Stress = %g psi\nTrue Strain at fracture = %g\nEngineering strain at fracture = %g')%(E_St,T_fr_St,e_f,e);

Engineering Stress at maximum load = 99852.1 psi
True Fracture Stress = 112785 psi
True Strain at fracture = 0.344939
Engineering strain at fracture = 0.411903


### Example 3.2, Yielding Criteria for Ductile Metals, Page No. 78¶

In [2]:
from math import sqrt

#variable declaration
sigma00=500;
sigma_z=-50;
sigma_y=100;
sigma_x=200;
T_xy=30;
T_yz=0;
T_xz=0;

#calculation
sigma0=sqrt((sigma_x-sigma_y)**2+(sigma_y-sigma_z)**2+(sigma_z-sigma_x)**2+6*(T_xy**2+T_yz**2+T_xz**2))/sqrt(2);
s=sigma00/sigma0;

#result
print('\nSince the calculated value of sigma0 = %g MPa, which is less than the yield strength of the aluminium alloy\nThus safety factor is = %g')%(sigma0,s);

Since the calculated value of sigma0 = 224.054 MPa, which is less than the yield strength of the aluminium alloy
Thus safety factor is = 2.23161


### Example 3.3, Tresca Criterion, Page No. 81¶

In [6]:
#variable declaration
sigma00=500;
sigma_z=-50;
sigma_y=100;
sigma_x=200;
T_xy=30;
T_yz=0;
T_xz=0;

#calculation
sigma0=sigma_x-sigma_z;
s=sigma00/sigma0;

#result
print('\nSince the calculated value of sigma0 = %g MPa, which is less than the yield strength of the aluminium alloy\nThus safety factor is = %g')%(sigma0,s);

Since the calculated value of sigma0 = 250 MPa, which is less than the yield strength of the aluminium alloy
Thus safety factor is = 2


### Example 3.4, Levy-Mises Equation, Page No. 91¶

In [4]:
from math import sqrt

#variable declaration
r_t=20;
p=1000;

#calculation
sigma1=p*r_t;
sigma1=sigma1/1000;                     #conversion to ksi
sigma=sqrt(3)*sigma1/2;
e=(sigma/25)**(1/0.25);
e1=sqrt(3)*e/2;

#result
print('\nPlastic Strain = %g')%(e1);

Plastic Strain = 0.199532