# Chapter 8:Transformation of stress and strain and Yield and Fracture criteria¶

## Example 8.1 page number 405¶

In :
#Given
import math
o = 22.5     #degrees , The angle of infetisimal wedge
A = 1        #mm2 The area of the element
A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB
A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC
S_1 = 3 #MN The stresses applying on the element
S_2 = 2 #MN
S_3 = 2 #MN
S_4 = 1 #MN
F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas
F_2 = S_2*A_ab
F_3 = S_3*A_bc
F_4 = S_4*A_bc
#sum of F_N = 0 equilibrim in normal direction

#sum of F_s = 0 equilibrim in tangential direction

Stress_Normal = N/A #Mpa - The stress action in normal direction on AB
Stress_tan = S/A    #Mpa - The stress action in tangential direction on AB
print "The stress action in normal direction on AB",round(Stress_Normal,2),"Mpa"
print "The stress action in tangential direction on AB",round(Stress_tan,2),"Mpa"

The stress action in normal direction on AB 1.29 Mpa
The stress action in tangential direction on AB 2.12 Mpa


## Example 8.2 page number 413¶

In :
#Given
o = -22.5 #degrees , The angle of infetisimal wedge
A = 1     #mm2 The area of the element
import math
from numpy import array
A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB
A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC
S_1 = 3.0 #MN The stresses applying on the element
S_2 = 2.0 #MN
S_3 = 2.0 #MN
S_4 = 1.0 #MN
#Caliculations

F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas
F_2 = S_2*A_ab
F_3 = S_3*A_bc
F_4 = S_4*A_bc
#sum of F_N = 0 equilibrim in normal direction

#sum of F_s = 0 equilibrim in tangential direction

Stress_Normal = N/A #Mpa - The stress action in normal direction on AB
Stress_tan = S/A    #Mpa - The stress action in tangential direction on AB
print "a) The stress action in normal direction on AB",round(Stress_Normal,2),"Mpa"
print "a) The stress action in tangential direction on AB",round(Stress_tan,2),"Mpa"

#Part- b

S_max = (S_4+S_1)/2 + (((((S_4-S_1)/2)**2) + S_3**2)**0.5)   #Mpa - The maximum stress
S_min = (S_4+S_1)/2.0 - (((((S_4-S_1/2))**2) + S_3**2)**0.5) #Mpa - The minumum stress
k = 0.5*math.atan(S_3/((S_1-S_4)/2))                         #radians The angle of principle axis
k_1 = math.degrees(k)
k_2 = k_1+90 #The principle plane angles
print "b) The principle stress ",round(S_max,1),"Mpa tension"
print "b) The principle stress ",round(S_min,2),"Mpa compression"
print "b) The principle plane angles are",round(k_1,0),",",round(k_2,0),"degrees"

#part-c
#The maximum shear stress case
t_xy = (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum shear stress case
K = 0.5*math.atan((-(S_1-S_4)/(2*S_3)))     #radians The angle of principle axis
K_0 = math.degrees(K)
if K_0<0:
K_1 = K_0+90
else:
K_1 = K_0
K_2 = K_1+90 #PRinciple plain angles
print "c) The maximum shear is ",round(T_xy,2),"Mpa"
S_mat_a = array([round(S_max,1),round(S_min,1),0])                       #MPa maximum stress matrix
S_mat_b = array([(S_4+S_1)/2,round(T_xy,2),round(T_xy,2),(S_4+S_1)/2])   #MPa maximum stress matrix at maximum shear
print "a)",S_mat_a,"Mpa"
print "b)",S_mat_b,"Mpa"

a) The stress action in normal direction on AB 4.12 Mpa
a) The stress action in tangential direction on AB 0.71 Mpa
b) The principle stress  4.2 Mpa tension
b) The principle stress  -0.06 Mpa compression
b) The principle plane angles are 32.0 , 122.0 degrees
c) The maximum shear is  -2.24 Mpa
a) [ 4.2 -0.1  0. ] Mpa
b) [ 2.   -2.24 -2.24  2.  ] Mpa


## Example 8.3 page number 421¶

In :
#Given
import math
S_x = -2 #Mpa _ the noraml stress in x direction
S_y = 4 #Mpa _ the noraml stress in Y direction
c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle
point_x = -2 #The x coordinate of a point on mohr circle
point_y = 4  #The y coordinate of a point on mohr circle
Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle
S_1  = Radius +1#MPa The principle stress
S_2 = -Radius +1 #Mpa The principle stress
S_xy_max = Radius #Mpa The maximum shear stress
print "The principle stresses are",S_1 ,"Mpa",S_2,"Mpa"
print "The maximum shear stress",S_xy_max,"Mpa"

The principle stresses are 6.0 Mpa -4.0 Mpa
The maximum shear stress 5.0 Mpa


## Example 8.4 page number 423¶

In :
#Given
import math
S_x = 3.0 #Mpa _ the noraml stress in x direction
S_y = 1.0 #Mpa _ the noraml stress in Y direction
c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle
point_x = 1 #The x coordinate of a point on mohr circle
point_y = 3  #The y coordinate of a point on mohr circle
#Caliculations

Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle
#22.5 degrees line is drawn
o = 22.5 #degrees
a = 71.5 - 2*o #Degrees, from diagram
stress_n = c + Radius*math.sin(math.degrees(o)) #Mpa The normal stress on the plane
stress_t =  Radius*math.cos(math.degrees(o)) #Mpa The tangential stress on the plane
print "The normal stress on the 221/2 plane ",round(stress_n,2),"Mpa"
print "The tangential stress on the 221/2 plane ",round(stress_t,2),"Mpa"

The normal stress on the 221/2 plane  4.82 Mpa
The tangential stress on the 221/2 plane  1.43 Mpa


## Example 8.7 page number 437¶

In :
import math
e_x = -500   #10-6 m/m The contraction in X direction
e_y = 300   #10-6 m/m The contraction in Y direction
e_xy = -600 #10-6 m/m discorted angle
centre = (e_x + e_y)/2  #10-6 m/m
point_x = -500 #The x coordinate of a point on mohr circle
point_y = 300  #The y coordinate of a point on mohr circle
Radius = 500   #10-6 m/m - from mohr circle
e_1  = Radius +centre    #MPa The principle strain
e_2 = -Radius +centre    #Mpa The principle strain
k = math.atan(300.0/900) # from geometry
k_1 = math.degrees(k)
print "The principle strains are",e_1,"um/m",e_2,"um/m"
print "The angle of principle plane",round(k_1,2) ,"degrees"

The principle strains are 400 um/m -600 um/m
The angle of principle plane 18.43 degrees


## Example 8.8 page number 441¶

In :
#Given
e_0 = -500 #10-6 m/m
e_45 = 200 #10-6 m/m
e_90 = 300 #10-6 m/m
E = 200    #Gpa - youngs modulus of steel
v = 0.3    # poissions ratio
#Caliculations

e_xy = 2*e_45 - (e_0 +e_90 ) #10-6 m/m from equation 8-40 in text
# from example 8.7
e_x = -500        #10-6 m/m The contraction in X direction
e_y = 300         #10-6 m/m The contraction in Y direction
e_xy = -600       #10-6 m/m discorted angle
centre = (e_x + e_y)/2  #10-6 m/m
point_x = -500          #The x coordinate of a point on mohr circle
point_y = 300           #The y coordinate of a point on mohr circle
Radius = 500            #10-6 m/m - from mohr circle
e_1  = Radius +centre #MPa The principle strain
e_2 = -Radius +centre #Mpa The principle strain

stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) #Mpa the stress in this direction
stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) #Mpa the stress in this direction
print"The principle stresses are ",round(stress_1,2),"Mpa",round(stress_2,2),"MPa"

The principle stresses are  48.35 Mpa -105.49 MPa